If $$tan^{-1}x+2 cot^{-1}x=\dfrac{2\pi}{3}$$, then x is equal to

$$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$

If $$\alpha,\beta \,\,and\,\, \gamma$$ are the roots of $$\displaystyle \tan^{-1} \left ( x-1 \right ) +\tan^{-1} x + \tan^{-1} \left ( x+1 \right ) = \tan^{-1} 3x$$, then

$$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{-1}{4}$$

If f(x) = $$(\sin^{-1}x)^2 +(\cos^{-1}x)^2$$, then.

f(x) has the least value of $$\dfrac{5\pi^2}{4}$$

f(x) has the least value of $$\dfrac{\pi^2}{8}$$

f(x) has the least value of $$\dfrac{5\pi^2}{8}$$

f(x) has the least value of $$\dfrac{\pi^2}{16}$$

If $$sin^{-1}x+sin^{-1}y=\dfrac{\pi}{2}$$ and $$\sin 2x=\cos 2y$$, then

$$y=-\dfrac{\pi}{12}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$$

$$x=\dfrac{\pi}{12}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$$

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 6

The value of

\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8}

$\tan^{-1}\dfrac{1}{3}+\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{7}+\tan^{-1}\dfrac{1}{8}$ is ___________.

0%
$\dfrac{11\pi}{5}$
0%
$\dfrac{\pi}{4}$
0%
$\pi$
0%
$\dfrac{3\pi}{4}$

Explanation

\tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{5}) + \tan^{- 1} (\frac{1}{7}) + \tan^{- 1} (\frac{1}{8})

$\tan^{-1} \left (\dfrac{1}{3} \right ) + \tan^{-1} \left (\dfrac{1}{5} \right ) + \tan^{-1} \left (\dfrac{1}{7} \right ) + \tan^{-1} \left (\dfrac{1}{8} \right )$

= \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{15}}) + \tan^{- 1} (\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{56}})

$= \tan^{-1} \left (\dfrac{\dfrac{1}{3} + \dfrac{1}{5}}{1 - \dfrac{1}{15}} \right ) + \tan^{-1} \left (\dfrac{\dfrac{1}{7} + \dfrac{1}{8}}{1 - \dfrac{1}{56}} \right )$

= \tan^{- 1} (\frac{\frac{8}{15}}{\frac{14}{15}}) + \tan^{- 1} (\frac{\frac{15}{56}}{\frac{55}{56}})

$= \tan^{-1} \left (\dfrac{\dfrac{8}{15}}{\dfrac{14}{15}} \right ) + \tan^{-1} \left (\dfrac{\dfrac{15}{56}}{\dfrac{55}{56}} \right )$

= \tan^{- 1} (\frac{8}{14}) + \tan^{- 1} (\frac{15}{55})

$= \tan^{-1} \left (\dfrac{8}{14} \right ) + \tan^{-1} \left (\dfrac{15}{55} \right )$

= \tan^{- 1} (\frac{4}{7}) + \tan^{- 1} (\frac{3}{11})

$= \tan^{-1} \left (\dfrac{4}{7} \right ) + \tan^{-1} \left (\dfrac{3}{11} \right )$

= \tan^{- 1} (\frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{12}{77}})

$= \tan^{-1} \left (\dfrac{\dfrac{4}{7} + \dfrac{3}{11}}{1 - \dfrac{12}{77}} \right )$

= \tan^{- 1} (\frac{\frac{65}{77}}{\frac{65}{77}})

$= \tan^{-1} \left (\dfrac{\dfrac{65}{77}}{\dfrac{65}{77}} \right )$

= \tan^{- 1} (1)

$= \tan^{-1} (1)$

= \tan^{- 1} (\tan \frac{π}{4})

$= \tan^{-1} \left (\tan \dfrac{\pi}{4} \right )$

= \frac{π}{4}

$= \dfrac{\pi}{4}$

\tan [2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}] =

$\tan \left [2\tan^{-1}\dfrac {1}{5}-\dfrac {\pi}{4}\right]=$ ?

0%
$\dfrac {7}{17}$
0%
$\dfrac {-7}{17}$
0%
$\dfrac {7}{12}$
0%
$\dfrac {-7}{12}$

Explanation

Let

2 \tan^{- 1} \frac{1}{5} = θ

$2\tan^{-1}\dfrac {1}{5}=\theta$ .

Then,

\tan^{- 1} \frac{1}{5} = \frac{1}{2} θ \Rightarrow \tan \frac{θ}{2} = \frac{1}{5}

$\tan^{-1}\dfrac {1}{5}=\dfrac {1}{2}\theta \Rightarrow \tan \dfrac {\theta}{2} =\dfrac {1}{5}$

∴ \tan θ = \frac{2 \tan \frac{1}{2} θ}{1 - \tan^{2} \frac{1}{2} θ} = \frac{(2 \times \frac{1}{5})}{(1 - \frac{1}{25})} = (\frac{2}{5} \times \frac{25}{24}) = \frac{5}{12}

$\therefore \ \tan \theta =\dfrac {2\tan \dfrac {1}{2}\theta}{1-\tan^2 \dfrac {1}{2}\theta}=\dfrac {\left(2\times \dfrac {1}{5}\right)}{\left(1-\dfrac {1}{25}\right)}=\left (\dfrac {2}{5}\times \dfrac {25}{24}\right)=\dfrac {5}{12}$

∴ \tan [2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}] = \tan (θ - \frac{π}{4}) = \frac{\tan θ - \tan \frac{π}{4}}{1 + \tan θ \cdot \tan \frac{π}{4}} = \frac{(\frac{5}{12} - 1)}{(1 + \frac{5}{12} \times 1)} = \frac{(\frac{- 7}{12})}{(\frac{17}{12})} = \frac{- 7}{17}

$\therefore \ \tan \left [2\tan^{-1}\dfrac {1}{5}-\dfrac {\pi}{4}\right]=\tan \left (\theta -\dfrac {\pi}{4}\right)=\dfrac {\tan \theta -\tan \dfrac {\pi}{4}}{1+\tan \theta \cdot \tan \dfrac {\pi}{4}}=\dfrac {\left (\dfrac {5}{12}-1\right)}{\left(1+\dfrac {5}{12}\times 1\right)}=\dfrac {\left (\dfrac {-7}{12}\right)}{\left (\dfrac {17}{12}\right)}=\dfrac {-7}{17}$

\sin (\cos^{- 1} \frac{3}{5}) =

$\sin \left (\cos^{-1}\dfrac {3}{5}\right)=$ ?

0%
$\dfrac {3}{4}$
0%
$\dfrac {4}{5}$
0%
$\dfrac {3}{5}$
0%
$none\ of\ these$

\cos (\tan^{- 1} \frac{3}{4}) =

$\cos \left(\tan^{-1} \dfrac {3}{4}\right)=$ ?

0%
$\dfrac {3}{5}$
0%
$\dfrac {4}{5}$
0%
$\dfrac {4}{9}$
0%
$none\ of\ these$

Explanation

Let

\tan^{- 1} \frac{3}{4} = x

$\tan^{-1}\dfrac {3}{4}=x$ , where

x \in (\frac{- π}{2}, \frac{π}{2})

$x\in \left(\dfrac {-\pi}{2}, \dfrac {\pi}{2}\right)$ .

∴ \tan x = \frac{3}{4}

$\therefore \ \tan x =\dfrac {3}{4}$ and since

x \in (\frac{- π}{2}, \frac{π}{2})

$x \in \left(\dfrac {-\pi}{2}, \dfrac {\pi}{2}\right)$ , where have

\cos x > 0

$\cos x > 0$

∴ \cos x = \frac{1}{\sec x} = \frac{1}{\sqrt{1 + \tan^{2} x}} = \frac{1}{\sqrt{1 + \frac{9}{16}}} = \frac{4}{5}

$\therefore \ \cos x =\dfrac {1}{\sec x}=\dfrac {1}{\sqrt {1+\tan^2 x}}=\dfrac {1}{\sqrt {1+\dfrac {9}{16}}}=\dfrac {4}{5}$

∴ \cos (\tan^{- 1} \frac{3}{4}) = \cos x = \frac{4}{5}

$\therefore \ \cos \left (\tan^{-1}\dfrac {3}{4}\right)=\cos x =\dfrac {4}{5}$ .

Evaluate :

\tan \frac{1}{2} (\cos^{- 1} \frac{\sqrt{5}}{3})

$\tan \dfrac {1}{2}\left (\cos^{-1}\dfrac {\sqrt 5}{3}\right)$

0%
$\dfrac {(3-\sqrt 5)}{2}$
0%
$\dfrac {(3+\sqrt 5)}{2}$
0%
$\dfrac {(5-\sqrt 3)}{2}$
0%
$\dfrac {(5+\sqrt 3)}{2}$

x \neq 0

$x\neq 0$ then

\cos (\tan^{- 1} x + \cot^{- 1} x) =

$\cos (\tan^{-1}x+\cot^{-1}x)=$ ?

0%
$-1$
0%
$1$
0%
$0$
0%
$none\ of\ these$

The value of

\sin (\sin^{- 1} \frac{1}{2} + \cos^{- 1} \frac{1}{2}) =

$\sin \left (\sin^{-1}\dfrac {1}{2}+\cos^{-1}\dfrac {1}{2}\right)=$ ?

0%
$0$
0%
$1$
0%
$-1$
0%
$none\ of\ these$

Explanation

As we know that,

\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}, x \in [- 1, 1]

$\sin^{-1} x +\cos^{-1}x=\dfrac {\pi}{2},\,x\in[-1,1]$

\sin (\sin^{- 1} \frac{1}{2} + \cos^{- 1} \frac{1}{2}) = \sin \frac{π}{2} = 1

$\sin \left (\sin^{-1}\dfrac {1}{2}+\cos^{-1}\dfrac {1}{2}\right)=\sin \dfrac {\pi}{2}=1$

The value of

\sin (\cos^{- 1} \frac{3}{5})

$\sin \left(\cos^{-1}\dfrac {3}{5}\right)$ is

0%
$\dfrac {2}{5}$
0%
$\dfrac {4}{5}$
0%
$\dfrac {-2}{5}$
0%
$none\ of\ these$

\sin [2 \tan^{- 1} \frac{5}{8}]

$\sin \left [2\tan^{-1}\dfrac {5}{8}\right]$

0%
$\dfrac {25}{64}$
0%
$\dfrac {80}{89}$
0%
$\dfrac {75}{128}$
0%
$none\ of\ these$

\sin [2 \sin^{- 1} \frac{4}{5}]

$\sin \left [2\sin^{-1}\dfrac{4}{5}\right]$

0%
$\dfrac {12}{25}$
0%
$\dfrac {16}{25}$
0%
$\dfrac {24}{25}$
0%
$none\ of\ these$

\sin {\frac{π}{3} - \sin^{- 1} (\frac{- 1}{2})} =

$\sin \left\{\dfrac {\pi}{3}-\sin^{-1}\left (\dfrac {-1}{2}\right) \right\}=$ ?

0%
$1$
0%
$0$
0%
$\dfrac {-1}{2}$
0%
$none\ of\ these$

\tan {\cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{2}{3}} =

$\tan \left\{\cos^{-1}\dfrac {4}{5}+\tan^{-1}\dfrac {2}{3}\right\}=$ ?

0%
$\dfrac {13}{6}$
0%
$\dfrac {17}{6}$
0%
$\dfrac {19}{6}$
0%
$\dfrac {23}{6}$

Explanation

\cos^{- 1} x = \tan^{- 1} \frac{\sqrt{1 - x^{2}}}{x} \Rightarrow \cos^{- 1} \frac{4}{5} = \tan^{- 1} \frac{\sqrt{1 - \frac{16}{25}}}{(\frac{4}{5})} = \tan^{- 1} \frac{3}{4}

$\cos ^{ -1 }{ x } =\tan ^{ -1 }{ \dfrac { \sqrt { 1-{ x }^{ 2 } } }{ x } } \Rightarrow \cos ^{ -1 }{ \dfrac { 4 }{ 5 } } =\tan ^{ -1 }{ \dfrac { \sqrt { 1-\dfrac { 16 }{ 25 } } }{ \left( \dfrac { 4 }{ 5 } \right) } } =\tan ^{ -1 }{ \dfrac { 3 }{ 4 } }$

∴ \cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{2}{3} = \tan^{- 1} \frac{3}{4} + \tan^{- 1} \frac{2}{3} = \tan^{- 1} (\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}) = \tan^{- 1} \frac{17}{6}

$\therefore \cos ^{ -1 }{ \dfrac { 4 }{ 5 } } +\tan ^{ -1 }{ \dfrac { 2 }{ 3 } } =\tan ^{ -1 }{ \dfrac { 3 }{ 4 } } +\tan ^{ -1 }{ \dfrac { 2 }{ 3 } } =\tan ^{ -1 }{ \left( \dfrac { \dfrac { 3 }{ 4 } +\dfrac { 2 }{ 3 } }{ 1-\dfrac { 3 }{ 4 } \times \dfrac { 2 }{ 3 } } \right) } =\tan ^{ -1 }{ \dfrac { 17 }{ 6 } }$

∴

$\therefore$ given exp.

= \tan {\tan^{- 1} \frac{17}{6}} = \frac{17}{6}

$=\tan { \left\{ \tan ^{ -1 }{ \dfrac { 17 }{ 6 } } \right\} } =\dfrac { 17 }{ 6 }$

Evaluate :

\cot (\tan^{- 1} x + \cot^{- 1} x)

$\cot (\tan^{-1}x+\cot^{-1}x)$

0%
$1$
0%
$\dfrac {1}{2}$
0%
$0$
0%
$none\ of\ these$

Evaluate :

\cos (2 \tan^{- 1} \frac{1}{2})

$\cos \left(2\tan^{-1}\dfrac {1}{2}\right)$

0%
$\dfrac {3}{5}$
0%
$\dfrac {4}{5}$
0%
$\dfrac {7}{8}$
0%
$none\ of\ these$

The value of

s i n^{- 1} {(s i n π / 3) \frac{x}{\sqrt{x^{2} + k^{2} - k x}}} - c o s^{- 1} {c o s π / 6 \frac{x}{\sqrt{x^{2} + k^{2} - k x}}}

$\displaystyle sin^{-1} \left \{ (sin\,\pi/3) \dfrac{x}{\sqrt{x^2 + k^2 - kx}} \right \} - cos^{-1} \left \{ cos\pi /6 \,\dfrac{x}{\sqrt{x^2 + k^2 - kx}} \right \}$

(w h e r e \frac{k}{2} < x < 2 k, k > 0)

$\left (where \,\dfrac{k}{2} < x < 2k , \,k > 0 \right )$ is

0%
$tan^{-1} \left ( \dfrac{2x^2 + xk - k^2}{x^2 - 2xk + k^2} \right )$
0%
$tan^{-1} \left ( \dfrac{x^2 + 2xk - k^2}{x^2 - 2xk + k^2} \right )$
0%
$tan^{-1} \left ( \dfrac{x^2 + 2xk - 2k^2}{2x^2 - 2xk + 2k^2} \right )$
0%
none of these

s i n^{- 1} \frac{5}{x} + s i n^{- 1} \frac{12}{x} = \frac{π}{2}

$sin^{-1}\dfrac{5}{x}+sin^{-1}\dfrac{12}{x}=\dfrac{\pi}{2}$ , then x is equal to

0%
$\dfrac{7}{13}$
0%
$\dfrac{4}{3}$
0%
$13$
0%
$\dfrac{13}{7}$

Explanation

Put

s i n^{- 1} \frac{5}{x} = A ⟹ \frac{5}{x} = \sin A

$sin^{-1}\dfrac{5}{x}=A\implies \dfrac{5}{x}=\sin A$

\sin^{- 1} \frac{12}{x} = B ⟹ \frac{12}{x} = \sin B ⟹ A + B = \frac{π}{2}

$\sin^{-1}\dfrac{12}{x}=B\implies \dfrac{12}{x}=\sin B \implies A+B=\dfrac{\pi}{2}$

⟹ \sin A = \sin (\frac{π}{2} - B) = \cos B = \sqrt{1 - s i n^{2} B}

$\implies\sin A=\sin(\dfrac{\pi}{2}-B)=\cos B=\sqrt{1-sin^{2}B}$

⟹ \frac{5}{x} = \sqrt{1 - \frac{144}{x^{2}}} ⟹ \frac{169}{x^{2}} = 1

$\implies\dfrac{5}{x}=\sqrt{1-\dfrac{144}{x^2}}\implies \dfrac{169}{x^2}=1$

⟹ x^{2} = 169 ⟹ x = 13

$\implies x^2=169 \implies x=13$ [

x = - 13

$x = -13$ does not satisfy the given equation]

\tan^{1} x + \tan^{1} y + \tan^{1} z = \frac{π}{2}

$\tan^{1}x + \tan^{1}y + \tan^{1}z = \dfrac {\pi}{2}$ , then

0%
$xy+yz+zx-xyz=0$
0%
$xy+yz+zx+xyz=0$
0%
$xy+yz+zx+1=0$
0%
$xy+yz+zx-1=0$

Explanation

G i v e n t h a t \tan^{1} x + \tan^{1} y + \tan^{1} z = \frac{π}{2}

$Given\,\,that \tan^{1}x + \tan^{1}y + \tan^{1}z = \dfrac {\pi}{2}$

⟹ \tan^{- 1} [\frac{x + y + z - x y z}{1 - x y - y z - z x}] = \frac{π}{2}

$\implies \tan^{-1}\Bigg[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\Bigg]=\dfrac{\pi}{2}$

For this denominator of

[\frac{x + y + z - x y z}{1 - x y - y z - z x}]

$\Bigg[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\Bigg]$ must be zero.

hence

x y + y z + z x - 1 = 0

$xy+yz+zx-1=0$

$cos^{-1}(cos(\dfrac{5\pi}{4}))$ is given by

0%
$\dfrac{5\pi}{4}$
0%
$\dfrac{3\pi}{4}$
0%
$\dfrac{-\pi}{4}$
0%
none of these

Explanation

c o s^{- 1} (c o s \frac{5 π}{4}) = c o s^{- 1} (c o s (2 π - \frac{5 π}{4})) = c o s^{- 1} (c o s (\frac{3 π}{4})) = \frac{3 π}{4}

$cos^{-1}\bigg(cos\dfrac{5\pi}{4}\bigg)=cos^{-1}\Bigg(cos\bigg(2\pi-\dfrac{5\pi}{4}\bigg)\Bigg)=cos^{-1}\Bigg(cos\bigg(\dfrac{3\pi}{4}\bigg)\Bigg)=\dfrac{3\pi}{4}$

The value of $sin^{-1}\Bigg(cot\bigg(sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+cos^{-1}\dfrac{\sqrt{12}}{4}+\sec^{-1}\sqrt{2}\bigg)\Bigg)$ is

0%
$0$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{3}$
0%
none of these

Explanation

We have

s i n^{- 1} (c o t (s i n^{- 1} \sqrt{\frac{2 - \sqrt{3}}{4}} + c o s^{- 1} \frac{\sqrt{12}}{4} + \sec^{- 1} \sqrt{2}))

$sin^{-1}\Bigg(cot\bigg(sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+cos^{-1}\dfrac{\sqrt{12}}{4}+\sec^{-1}\sqrt{2}\bigg)\Bigg)$

\sin^{- 1} (\cot (s i n^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + c o s^{- 1} \frac{\sqrt{3}}{2} + \cos^{- 1} \frac{1}{\sqrt{2}}))

$\sin^{-1}\Bigg(\cot\bigg(sin^{-1}\bigg(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\bigg)+cos^{-1}\dfrac{\sqrt{3}}{2}+\cos^{-1}\dfrac{1}{\sqrt{2}}\bigg)\Bigg)$

\sin^{- 1} [\cot (15^{\circ} + 30^{\circ} + 45^{\circ})]

$\sin^{-1}[\cot(15^{\circ}+30^{\circ}+45^{\circ})]$

\sin^{- 1} [\cot (90^{\circ})]

$\sin^{-1}[\cot(90^{\circ})]$

\sin^{- 1} (0) = 0

$\sin^{-1}(0)=0$

\tan^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} \tan^{- 1} x

$\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{1}{2}\tan^{-1}x$ , then x is equal to

0%
$1$
0%
$\sqrt{3}$
0%
$\dfrac{1}{\sqrt{3}}$
0%
none of these

Explanation

We have

\tan^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} \tan^{- 1} x

$\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{1}{2}\tan^{-1}x$

⟹ \tan^{- 1} [\frac{1 - \tan θ}{1 + \tan θ}] = \frac{1}{2} θ

$\implies\,\tan^{-1}\Bigg[\dfrac{1-\tan\theta}{1+\tan\theta}\Bigg]=\dfrac{1}{2}\theta$

⟹ \tan^{- 1} [\frac{\tan \frac{π}{4} - \tan θ}{1 + \tan \frac{π}{4} \tan θ}] = \frac{1}{2} θ

$\implies\,\tan^{-1}\Bigg[\dfrac{\tan\dfrac{\pi}{4}-\tan\theta}{1+\tan\dfrac{\pi}{4}\tan\theta}\Bigg]=\dfrac{1}{2}\theta$

⟹ \tan^{- 1} \tan (\frac{π}{4} - θ) = \frac{θ}{2}

$\implies\,\tan^{-1}\tan\bigg(\dfrac{\pi}{4}-\theta\bigg)=\dfrac{\theta}{2}$

⟹ \frac{π}{4} - θ = \frac{θ}{2}

$\implies\dfrac{\pi}{4}-\theta=\dfrac{\theta}{2}$

⟹ θ = \frac{π}{6} = \tan^{- 1} x

$\implies\theta=\dfrac{\pi}{6}=\tan^{-1}x$

⟹ x = t a n \frac{π}{6} = \frac{1}{\sqrt{3}}

$\implies x=tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}$

t a n^{- 1} x + 2 c o t^{- 1} x = \frac{2 π}{3}

$tan^{-1}x+2 cot^{-1}x=\dfrac{2\pi}{3}$ , then x is equal to

0%
$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$
0%
$3$
0%
$\sqrt{3}$
0%
$\sqrt{2}$

Explanation

\tan^{- 1} x + 2 \cot^{- 1} x = \frac{2 π}{3}

$\tan^{-1}x+2 \cot^{-1}x=\dfrac{2\pi}{3}$

⟹ t a n^{- 1} x = 2 (\frac{π}{3} - \cot^{- 1} x) = 2 (\frac{π}{3} - (\frac{π}{2} - \tan^{- 1} x)) = 2 (- \frac{π}{6} + \tan^{- 1} x)

$\implies tan^{-1}x=2\bigg(\dfrac{\pi}{3}-\cot^{-1}x\bigg)=2\bigg(\dfrac{\pi}{3}-(\dfrac{\pi}{2}-\tan^{-1}x)\bigg)=2\bigg(-\dfrac{\pi}{6}+\tan^{-1}x\bigg)$

⟹ \tan^{- 1} x = \frac{π}{3} ⟹ x = \tan \frac{π}{3} = \sqrt{3}

$\implies\tan^{-1}x=\dfrac{\pi}{3} \implies x=\tan\dfrac{\pi}{3}=\sqrt{3}$

3 \sin^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 t a n^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}

$3\sin^{-1}\bigg(\dfrac{2x}{1+x^2}\bigg)-4\cos^{-1}\bigg(\dfrac{1-x^2}{1+x^2}\bigg)+2tan^{-1}\bigg(\dfrac{2x}{1-x^2}\bigg)=\dfrac{\pi}{3}$ , where

| x | < 1

$|x|<1$ . then x is equal to

0%
$\dfrac{1}{\sqrt{3}}$
0%
$-\dfrac{1}{\sqrt{3}}$
0%
${\sqrt{3}}$
0%
$-\dfrac{\sqrt{3}}{4}$
0%
$\dfrac{\sqrt{3}}{2}$

Explanation

Put

x = \tan θ

$x=\tan\theta$

∴ 3 \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) - 4 \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) + 2 \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) = \frac{π}{3}

$\therefore\,3\sin^{-1}\bigg(\dfrac{2\tan\theta}{1+\tan^2\theta}\bigg)-4\cos^{-1}\bigg(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\bigg)+2\tan^{-1}\bigg(\dfrac{2\,\tan\theta}{1-\tan^2\theta}\bigg)=\dfrac{\pi}{3}$

⟹ 3 \sin^{- 1} (\sin 2 θ) - 4 \cos^{- 1} (\cos 2 θ) + 2 \tan^{- 1} (\tan 2 θ) = \frac{π}{3}

$\implies\,3\sin^{-1}(\sin2\theta)-4\cos^{-1}(\cos2\theta)+2\tan^{-1}(\tan2\theta)=\dfrac{\pi}{3}$

⟹ 3 (2 θ) - 4 (2 θ) + 2 (2 θ) = \frac{π}{3} ⟹ 2 θ = \frac{π}{3} ⟹ θ = \frac{π}{6} ⟹ x = \tan \frac{π}{6} = \frac{1}{\sqrt{3}}

$\implies\,3(2\theta)-4(2\theta)+2(2\theta)=\dfrac{\pi}{3}\,\implies2\theta=\dfrac{\pi}{3}\implies\theta=\dfrac{\pi}{6}\implies\,x=\tan \dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}$

The principal value of

\sin^{- 1} (\sin 10)

$\sin^{-1}(\sin 10)$ is

0%
$10$
0%
$10-3\pi$
0%
$3\pi-10$
0%
none of these

Explanation

\sin^{- 1} (\sin 10) = \sin^{- 1} [\sin (3 π - 10)] = 3 π - 10

$\sin^{-1}(\sin 10)=\sin^{-1}[\sin (3\pi-10)]=3\pi-10$

3 \tan^{- 1} (\frac{1}{2 + \sqrt{3}}) - \tan^{- 1} \frac{1}{x} = \frac{1}{3}

$3\tan^{-1}(\dfrac{1}{2+\sqrt{3}})-\tan^{-1}\dfrac{1}{x}=\dfrac{1}{3}$ , then x is equal to

0%
$1$
0%
$2$
0%
$3$
0%
$\sqrt{2}$

Explanation

The given equation can be written as

3 \tan^{- 1} (\frac{1}{2 + \sqrt{3}}) = \frac{1}{3} + \tan^{- 1} \frac{1}{x}

$3\tan^{-1}(\dfrac{1}{2+\sqrt{3}})=\dfrac{1}{3}+\tan^{-1}\dfrac{1}{x}$

⟹ 3 (15^{\circ}) = \tan^{- 1} \frac{\frac{1}{x} + \frac{1}{3}}{1 - \frac{1}{x} \frac{1}{3}} ⟹ 1 = \frac{3 + x}{3 x - 1} ⟹ x = 2

$\implies 3(15^{\circ})=\tan^{-1}\dfrac{\dfrac{1}{x}+\dfrac{1}{3}}{1-\dfrac{1}{x}\dfrac{1}{3}}\,\implies1=\dfrac{3+x}{3x-1}\implies x=2$

The value

2 t a n^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}]

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ is equal to

0%
$\cos^{-1}\Bigg[\dfrac{a\cos\theta+b}{b\cos\theta+a}\Bigg]$
0%
$\cos^{-1}\Bigg[\dfrac{b\cos\theta+a}{a\cos\theta+b}\Bigg]$
0%
$\cos^{-1}\Bigg[\dfrac{a\cos\theta}{b\cos\theta+a}\Bigg]$
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$\cos^{-1}\Bigg[\dfrac{b\cos\theta}{a\cos\theta+b}\Bigg]$

Explanation

2 t a n^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}]

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ =

c o s^{- 1} [\frac{1 - (\frac{a - b}{a + b}) \tan^{2} \frac{θ}{2}}{1 + (\frac{a - b}{a + b}) \tan^{2} \frac{θ}{2}}]

$cos^{-1}\Bigg[\dfrac{1-\bigg(\dfrac{a-b}{a+b}\bigg)\tan^2\dfrac{\theta}{2}}{1+\bigg(\dfrac{a-b}{a+b}\bigg)\tan^2\dfrac{\theta}{2}}\Bigg]$

2 t a n^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}]

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ =

c o s^{- 1} [\frac{(a + b) - (a - b) \tan^{2} \frac{θ}{2}}{(a + b) + a - b \tan^{2} \frac{θ}{2}}]

$cos^{-1}\Bigg[\dfrac{(a+b)-(a-b)\tan^2\dfrac{\theta}{2}}{(a+b)+{a-b}\tan^2\dfrac{\theta}{2}}\Bigg]$

2 t a n^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}]

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ =

c o s^{- 1} [\frac{a (1 - \tan^{2} \frac{θ}{2}) + b (1 + \tan^{2} \frac{θ}{2})}{a (1 + \tan^{2} \frac{θ}{2}) + b (1 - \tan^{2} \frac{θ}{2})}]

$cos^{-1}\Bigg[\dfrac{a(1-\tan^2\dfrac{\theta}{2})+b(1+\tan^2\dfrac{\theta}{2})}{a(1+\tan^2\dfrac{\theta}{2})+b(1-\tan^2\dfrac{\theta}{2})}\Bigg]$

2 t a n^{- 1} [\sqrt{\frac{a - b}{a + b}} \tan \frac{θ}{2}]

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ =

c o s^{- 1} [\frac{\frac{a (1 - \tan^{2} \frac{θ}{2})}{(1 + \tan^{2} \frac{θ}{2})} + b}{a + b \frac{(1 - \tan^{2} \frac{θ}{2})}{(1 + \tan^{2} \frac{θ}{2})}}]

$cos^{-1}\Bigg[\dfrac{\dfrac{a(1-\tan^2\dfrac{\theta}{2})}{(1+\tan^2\dfrac{\theta}{2})}+b}{a+b\dfrac{(1-\tan^2\dfrac{\theta}{2})}{(1+\tan^2\dfrac{\theta}{2})}}\Bigg]$

= \cos^{- 1} [\frac{a \cos θ + b}{b \cos θ + a}]

$=\cos^{-1}\Bigg[\dfrac{a\cos\theta+b}{b\cos\theta+a}\Bigg]$

x = \sec^{- 1} (x + \frac{1}{x}) + \sec^{- 1} (y + \frac{1}{y})

$x=\sec^{-1}\Bigg(x+\dfrac{1}{x}\Bigg)+\sec^{-1}\Bigg(y+\dfrac{1}{y}\Bigg)$ where xy < 0, then the possible values of z is (are)

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$\dfrac{8\pi}{10}$
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$\dfrac{7\pi}{10}$
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$\dfrac{9\pi}{10}$
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$\dfrac{21\pi}{20}$

Explanation

x y < 0 \Rightarrow x + \frac{1}{x} \geq 2, y + \frac{1}{y} \leq - 2

$xy<0\Rightarrow x+\dfrac{1}{x}\geq2,y+\dfrac{1}{y}\leq-2$
or

x + \frac{1}{x} \leq - 2, y + \frac{1}{y} \geq 2

$x+\dfrac{1}{x}\leq-2,y+\dfrac{1}{y}\geq2$

x + \frac{1}{x} \geq 2 \Rightarrow \sec^{- 1} (x + \frac{1}{x}) \in [\frac{π}{3}, \frac{π}{3})

$x+\dfrac{1}{x}\geq2\Rightarrow\sec^{-1}\Bigg(x+\dfrac{1}{x}\Bigg)\in[\dfrac{\pi}{3},\dfrac{\pi}{3})$

y + \frac{1}{y} \leq - 2 \Rightarrow \sec^{- 1} (y + \frac{1}{y}) \in (\frac{π}{2}, \frac{2 π}{3}] \Rightarrow z \in (\frac{5 π}{6}, \frac{7 π}{6})

$y+\dfrac{1}{y}\leq-2\Rightarrow\sec^{-1}\Bigg(y+\dfrac{1}{y}\Bigg)\in(\dfrac{\pi}{2},\dfrac{2\pi}{3}] \,\Rightarrow z\in\bigg(\dfrac{5\pi}{6},\dfrac{7\pi}{6}\bigg)$

t a n^{- 1} (\sin^{2} θ - 2 s i n θ + 3) + \cot^{- 1} (5^{\sec^{2} y} + 1) = \frac{π}{2}

$tan^{-1}(\sin^2\theta-2sin\theta+3)+\cot^{-1}(5^{\sec^2y}+1)=\dfrac{\pi}{2}$ , then the value of

c o s^{2} θ - \sin θ

$cos^2\theta-\sin\theta$ is equal to

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$0$
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$-1$
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$1$
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none of these

Explanation

From the given equation

\sin^{2} θ - 2 s i n θ + 3 = 5^{\sec^{2} y} + 1

$\sin^2\theta-2sin\theta+3=5^{\sec^2y}+1$ , we get

(\sin θ - 1)^{2} + 2 = 5^{\sec^{2} y} + 1

$(\sin\theta-1)^2+2=5^{\sec^2y}+1$

L . H . S . \leq 6, R . H . S . \geq 6

$L.H.S.\leq6 , R.H.S.\geq6$

Possible solution is

s i n θ = - 1

$sin\theta=-1$ when L.H.S. = R.H.S.

\Rightarrow c o s^{2} θ = 0 \Rightarrow c o s^{2} θ - \sin θ = 1

$\Rightarrow cos^2\theta=0\,\Rightarrow cos^2\theta-\sin\theta=1$

If f(x)=

s i n^{- 1} (\frac{\sqrt{3}}{2} x - \frac{1}{2} \sqrt{1 - x^{2}}), - \frac{1}{2} \leq x \leq 1

$sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}x-\dfrac{1}{2}\sqrt{1-x^2}\Bigg),-\dfrac{1}{2}\leq x\leq1$ , then f(x) is equal to

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$\sin^{-1}\bigg(\dfrac{1}{2}\bigg)-sin^{-1}x$
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$\sin^{-1}x-\dfrac{\pi}{6}$
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$\sin^{-1}x+\dfrac{\pi}{6}$
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none of these

Explanation

Let

x = \sin θ

$x=\sin\theta$ where

\frac{- 1}{2} \leq x \leq 1 ⟹ \frac{- π}{6} \leq θ \leq \frac{π}{2}

$\dfrac{-1}{2}\leq x\leq1\implies\,\dfrac{-\pi}{6}\leq\theta\leq\dfrac{\pi}{2}$

Then

f (x) = s i n^{- 1} (\frac{\sqrt{3}}{2} x - \frac{1}{2} \sqrt{1 - x^{2}})

$f(x)=sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}x-\dfrac{1}{2}\sqrt{1-x^2}\Bigg)$

= s i n^{- 1} (\frac{\sqrt{3}}{2} \sin θ - \frac{1}{2} \cos θ)

$=sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}\sin\theta-\dfrac{1}{2}\cos\theta\Bigg)$

= s i n^{- 1} (\sin (θ - \frac{π}{6}))

$=sin^{-1}\Bigg(\sin\bigg(\theta-\dfrac{\pi}{6}\bigg)\Bigg)$

= θ - \frac{π}{6} = s i n^{- 1} x - \frac{π}{6}

$=\theta-\dfrac{\pi}{6}=sin^{-1}x-\dfrac{\pi}{6}$

c o t^{- 1} (\sqrt{\cos a}) - t a n^{- 1} (\sqrt{\cos a}) = x

$cot^{-1}(\sqrt{\cos a})- tan^{-1}(\sqrt{\cos a})=x$ , then

\sin x

$\sin x$ is

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$tan^2\dfrac{\alpha}{2}$
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$cot^2\dfrac{\alpha}{2}$
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$\tan \alpha$
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$cot\dfrac{\alpha}{2}$

Explanation

c o t^{- 1} (\sqrt{\cos a}) - t a n^{- 1} (\sqrt{\cos a}) = x

$cot^{-1}(\sqrt{\cos a})- tan^{-1}(\sqrt{\cos a})=x$

t a n^{- 1} (\frac{1}{\sqrt{\cos a}}) - t a n^{- 1} (\sqrt{\cos a}) = x

$tan^{-1}(\dfrac{1}{\sqrt{\cos a}})- tan^{-1}(\sqrt{\cos a})=x$

t a n^{- 1} \frac{\frac{1}{\sqrt{\cos a}} - \sqrt{\cos a}}{1 + \frac{1}{\sqrt{\cos a}} \sqrt{\cos a}} = x

$tan^{-1}\dfrac{\dfrac{1}{\sqrt{\cos a}}-\sqrt{\cos a}}{1+\dfrac{1}{\sqrt{\cos a}}\sqrt{\cos a}}=x$

t a n^{- 1} \frac{1 - \cos a}{2 \sqrt{\cos a}} = x

$tan^{-1}\dfrac{1-\cos a}{2\sqrt{\cos a}}=x$

\frac{1 - \cos a}{2 \sqrt{\cos a}} = \tan x

$\dfrac{1-\cos a}{2\sqrt{\cos a}}=\tan x$

\frac{2 \sqrt{\cos a}}{1 - \cos a} = \cot x

$\dfrac{2\sqrt{\cos a}}{1-\cos a}=\cot x$

\sin x = \frac{1 - \cos a}{1 + \cos a} = \frac{2 \sin^{2} \frac{α}{2}}{2 \cos^{2} \frac{α}{2}}

$\sin x=\dfrac{1-\cos a}{1+\cos a}=\dfrac{2\sin^2\dfrac{\alpha}{2}}{2\cos^2\dfrac{\alpha}{2}}$ =

t a n^{2} \frac{α}{2}

$tan^2\dfrac{\alpha}{2}$

2 t a n^{- 1} (- 2)

$2tan^{-1}(-2)$ is equal to

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$-\cos^{-1}\bigg(\dfrac{-3}{5}\bigg)$
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$-\pi+\cos^{-1}\bigg(\dfrac{3}{5}\bigg)$
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$-\dfrac{\pi}{2}+\tan^{-1}\bigg(\dfrac{-3}{4}\bigg)$
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$-\pi+\cot^{-1}\bigg(\dfrac{-3}{4}\bigg)$

Explanation

Let

t a n^{- 1} (- 2) = θ \Rightarrow \tan θ = - 2 \Rightarrow θ = (\frac{- π}{2}, 0)

$tan^{-1}(-2)=\theta\,\Rightarrow\,\tan\theta=-2\,\Rightarrow\,\theta=(\dfrac{-\pi}{2},0)$

\Rightarrow 2 θ = (- π, 0)

$\Rightarrow2\theta=(-\pi,0)$

\cos (- 2 θ) = \cos (2 θ) = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} = \frac{- 3}{5}

$\cos(-2\theta)=\cos(2\theta)=\dfrac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\dfrac{-3}{5}$

\Rightarrow - 2 θ = \cos^{- 1} \frac{- 3}{5} = π - \cos^{- 1} \frac{3}{5}

$\Rightarrow -2\theta=\cos^{-1}\dfrac{-3}{5}=\pi-\cos^{-1}\dfrac{3}{5}$

\Rightarrow 2 θ = - π + \cos^{- 1} \frac{3}{5} = - π + \tan^{- 1} \frac{4}{3} = - π + \cot^{- 1} \frac{3}{4} = - \frac{π}{2} - t a n^{- 1} \frac{3}{4}

$\Rightarrow 2\theta=-\pi+\cos^{-1}\dfrac{3}{5}=-\pi+\tan^{-1}\dfrac{4}{3}=-\pi+\cot^{-1}\dfrac{3}{4}=-\dfrac{\pi}{2}-tan^{-1}\dfrac{3}{4}$

= - \frac{π}{2} + t a n^{- 1} \frac{- 3}{4}

$=-\dfrac{\pi}{2}+tan^{-1}\dfrac{-3}{4}$

The value of

\underset{| x | \to \infty}{l i m} c o s (t a n^{- 1} (s i n (t a n^{- 1} x)))

$\underset{|x|\rightarrow\infty}{lim}cos(tan^{-1}(sin(tan^{-1}x)))$ is equal to

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$-1$
0%
$\sqrt{2}$
0%
$-\dfrac{1}{\sqrt{2}}$
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$\dfrac{1}{\sqrt{2}}$

Explanation

\underset{| x | \to \infty}{l i m} c o s (t a n^{- 1} (s i n (t a n^{- 1} x))) = c o s (t a n^{- 1} (s i n (t a n^{- 1} \infty)))

$\underset{|x|\rightarrow\infty}{lim}cos(tan^{-1}(sin(tan^{-1}x)))=cos(tan^{-1}(sin(tan^{-1}\infty)))$

= c o s (t a n^{- 1} (s i n (\frac{π}{2}))

$=cos(tan^{-1}(sin(\dfrac{\pi}{2}))$

= c o s (t a n^{- 1} (1))

$=cos(tan^{-1}(1))$

= c o s (\frac{π}{4}) = \frac{1}{\sqrt{2}}

$=cos(\dfrac{\pi}{4})=\dfrac{1}{\sqrt{2}}$

x \in (\frac{- π}{2}, \frac{π}{2})

$x\in\bigg(\dfrac{-\pi}{2},\dfrac{\pi}{2}\bigg)$ , then the value

\tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{3 \sin 2 x}{5 + 3 \cos 2 x})

$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\sin 2x}{5+3\cos 2x}\Bigg)$ is

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$\dfrac{x}{2}$
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$2x$
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$3x$
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$x$

Explanation

\tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{3 \sin 2 x}{5 + 3 \cos 2 x})

$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\sin 2x}{5+3\cos 2x}\Bigg)$ =

\tan^{- 1} (\frac{\tan x}{4})

$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)$ +

\tan^{- 1} (\frac{\frac{6 \tan x}{1 + t a n^{2} x}}{5 + 3 \frac{1 - t a n^{2} x}{1 + t a n^{2} x}})

$\tan^{-1}\Bigg(\dfrac{\dfrac{6\tan x}{1+tan^{2}x}}{5+3\dfrac{1-tan^{2}x}{1+tan^{2}x}}\Bigg)$

= \tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{6 \tan x}{8 + 2 \tan^{2} x})

$=\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{6\tan x}{8+2\tan^2x}\Bigg)$

= \tan^{- 1} (\frac{\tan x}{4}) + \tan^{- 1} (\frac{3 \tan x}{4 + \tan^{2} x})

$=\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\tan x}{4+\tan^2x}\Bigg)$

= \tan^{- 1} (\frac{\frac{\tan x}{4} + \frac{3 \tan x}{4 + \tan^{2} x}}{1 - \frac{3 \tan^{2} x}{4 (4 + \tan^{2} x)}})

$=\tan^{-1}\Bigg(\dfrac{\dfrac{\tan x}{4}+\dfrac{3\tan x}{4+\tan^2x}}{1-\dfrac{3\tan^2x}{4(4+\tan^2x)}}\Bigg)$

= \tan^{- 1} (\frac{16 \tan x + t a n^{3} x}{16 + t a n^{2} x})

$=\tan^{-1}\Bigg(\dfrac{16\tan x+tan^{3}x}{16+tan^2x}\Bigg)$

= \tan^{- 1} (\tan x) = x

$=\tan^{-1}(\tan x)=x$

α, β a n d γ

$\alpha,\beta \,\,and\,\, \gamma$ are the roots of

\tan^{- 1} (x - 1) + \tan^{- 1} x + \tan^{- 1} (x + 1) = \tan^{- 1} 3 x

$\displaystyle \tan^{-1} \left ( x-1 \right ) +\tan^{-1} x + \tan^{-1} \left ( x+1 \right ) = \tan^{-1} 3x$ , then

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$\alpha+\beta+ \gamma=0$
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$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{-1}{4}$
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$\alpha\beta\gamma=1$
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$|\alpha-\beta|_{max}=1$

Explanation

Given,

\tan^{- 1} (x - 1) + \tan^{- 1} x + \tan^{- 1} (x + 1) = \tan^{- 1} 3 x

$\displaystyle \tan^{-1} \left ( x-1 \right ) +\tan^{-1} x + \tan^{-1} \left ( x+1 \right ) = \tan^{-1} 3x$

\Rightarrow \tan^{- 1} (x - 1) + \tan^{- 1} x = \tan^{- 1} 3 x - \tan^{- 1} (x + 1)

$\Rightarrow\displaystyle \tan^{-1} \left ( x-1 \right ) +\tan^{-1} x = \tan^{-1} 3x - \tan^{-1} \left ( x+1 \right )$

\Rightarrow \tan^{- 1} [\frac{(x - 1) + x}{1 - (x - 1) x}] = \tan^{- 1} [\frac{3 x - (x + 1)}{1 + (x + 1) 3 x}]

$\Rightarrow\tan^{-1}\Bigg[\dfrac{(x-1)+x}{1-(x-1)x}\Bigg]=\tan^{-1}\Bigg[\dfrac{3x-(x+1)}{1+(x+1)3x}\Bigg]$

\Rightarrow [\frac{2 x - 1}{1 - x^{2} + x}] = [\frac{2 x - 1}{1 + 3 x^{2} + 3 x}]

$\Rightarrow\Bigg[\dfrac{2x-1}{1-x^2+x}\Bigg]=\Bigg[\dfrac{2x-1}{1+3x^2+3x}\Bigg]$

\Rightarrow (1 - x^{2} + x) (2 x - 1) = (1 + 3 x^{2} + 3 x) (2 x - 1)

$\Rightarrow(1 - x^2 + x) (2x - 1) = ( 1 + 3x^2 + 3x) (2x- 1)$

\Rightarrow x = 0, \pm \frac{1}{2}

$\Rightarrow x=0,\pm \dfrac{1}{2}$

Given that

α, β a n d γ

$\alpha,\beta \,\,and\,\, \gamma$ are the roots of given equation.

α + β + γ = \frac{1}{2} - \frac{1}{2} + 0 = 0

$\alpha+\beta+ \gamma=\dfrac 12-\dfrac12+0=0$

α β + β γ + γ α = 0 + (\frac{1}{2} \times \frac{- 1}{2}) + 0 = \frac{- 1}{4}

$\alpha\beta+\beta\gamma+\gamma\alpha=0+\left(\dfrac{1}{2}\times \dfrac{-1}{2}\right )+0=\dfrac{-1}{4}$

α β γ = 0

$\alpha\beta\gamma=0$

| α - β |_{m a x} = | \frac{- 1}{2} - \frac{1}{2} | = 1

$|\alpha-\beta|_{max}=\left|\dfrac{-1}2-\dfrac 12\right|=1$

If f(x) =

(\sin^{- 1} x)^{2} + (\cos^{- 1} x)^{2}

$(\sin^{-1}x)^2 +(\cos^{-1}x)^2$ , then.

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f(x) has the least value of $\dfrac{\pi^2}{8}$
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f(x) has the least value of $\dfrac{5\pi^2}{8}$
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f(x) has the least value of $\dfrac{\pi^2}{16}$
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f(x) has the least value of $\dfrac{5\pi^2}{4}$

Explanation

Let

f (x) = (\sin^{- 1} x)^{2} + (\cos^{- 1} x)^{2}

$f (x) = (\sin^{-1}x)^2 +(\cos^{-1}x)^2$ ,

= (\sin^{- 1} x + \cos^{- 1} x)^{2} - 2 \sin^{- 1} x \cos^{- 1} x

$=(\sin^{-1}x+\cos^{-1}x)^2 - 2\sin^{-1}x\cos^{-1}x$

= \frac{π^{2}}{4} - 2 \sin^{- 1} x [\frac{π}{2} - \sin^{- 1} x]

$=\dfrac{\pi^2}{4}- 2\sin^{-1}x\Bigg[\dfrac{\pi}{2}-\sin^{-1}x\Bigg]$

= \frac{π^{2}}{4} - π \sin^{- 1} x 2 (\sin^{- 1} x)^{2}

$=\dfrac{\pi^2}{4}- \pi\sin^{-1}x2(\sin^{-1}x)^2$

= 2 (\sin^{- 1} x - \frac{π}{4})^{2} + 2 [\frac{π^{2}}{16}]

$=2\Bigg(\sin^{-1}x-\dfrac{\pi}{4}\Bigg)^2+2\Bigg[\dfrac{\pi^2}{16}\Bigg]$

Now,

- \frac{π}{2} \leq s i n^{- 1} x \leq \frac{π}{2}

$-\dfrac{\pi}{2}\leq sin^{-1}x\leq\dfrac{\pi}{2}$

\Rightarrow - \frac{3 π}{4} \leq s i n^{- 1} x - \frac{π}{4} \leq \frac{π}{4}

$\Rightarrow-\dfrac{3\pi}{4}\leq sin^{-1}x-\dfrac{\pi}{4}\leq\dfrac{\pi}{4}$

\Rightarrow 0 \leq (s i n^{- 1} x - \frac{π}{4})^{2} \leq \frac{9 π^{2}}{16}

$\Rightarrow0\leq \bigg(sin^{-1}x-\dfrac{\pi}{4}\bigg)^2\leq\dfrac{9\pi^2}{16}$

\Rightarrow 0 \leq 2 (s i n^{- 1} x - \frac{π}{4})^{2} \leq \frac{9 π^{2}}{8}

$\Rightarrow0\leq 2\bigg(sin^{-1}x-\dfrac{\pi}{4}\bigg)^2\leq\dfrac{9\pi^2}{8}$

\Rightarrow \frac{π^{2}}{8} \leq 2 (s i n^{- 1} x - \frac{π}{4})^{2} + \frac{π^{2}}{8} \leq \frac{5 π^{2}}{4}

$\Rightarrow\dfrac{\pi^2}{8}\leq 2\bigg(sin^{-1}x-\dfrac{\pi}{4}\bigg)^2+\dfrac{\pi^2}{8}\leq\dfrac{5\pi^2}{4}$

s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}

$sin^{-1}x+sin^{-1}y=\dfrac{\pi}{2}$ and

\sin 2 x = \cos 2 y

$\sin 2x=\cos 2y$ , then

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$x=\dfrac{\pi}{8}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$
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$y=-\dfrac{\pi}{12}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$
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$x=\dfrac{\pi}{12}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$
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$y=-\dfrac{\pi}{8}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$

Explanation

For the given equation

0 \leq x, y \leq 1

$0\leq x,y\leq1$

Also,

\sin^{- 1} x + \sin^{- 1} y = \frac{π}{2}

$\sin^{-1}x+\sin^{-1}y=\dfrac{\pi}{2}$

\Rightarrow \sin^{- 1} x = \cos^{- 1} y = \sin^{- 1} \sqrt{1 - y^{2}}

$\Rightarrow \sin^{-1}x=\cos^{-1}y=\sin^{-1}\sqrt{1-y^2}$

\Rightarrow x = \sqrt{1 - y^{2}} \Rightarrow x^{2} + y^{2} = 1

$\Rightarrow x=\sqrt{1-y^2} \,\Rightarrow\, x^2+y^2=1$ (i)

Again,

\sin 2 x = \cos 2 y

$\sin 2x = \cos 2y$

\Rightarrow \cos (\frac{π}{2} - 2 x) = \cos 2 y

$\Rightarrow\cos\bigg(\dfrac{\pi}{2}-2x\bigg)=\cos 2y$

\frac{π}{2} - 2 x = 2 n y \pm 2 y, w h e r e n \in I

$\dfrac{\pi}{2}-2x=2ny\pm2y,where\,n \in I$

x \pm y = \frac{π}{4} - n π

$x\pm y=\dfrac{\pi}{4}-n\pi$ (ii)

From Eqs. (i) and (ii), we get

x = \frac{π}{8} + \sqrt{\frac{1}{2} - \frac{π^{2}}{64}}

$x=\dfrac{\pi}{8}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$ and

y = - \frac{π}{8} + \sqrt{\frac{1}{2} - \frac{π^{2}}{64}}

$y=-\dfrac{\pi}{8}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$

c o s^{- 1} x + c o s^{- 1} [\frac{x}{2} + \frac{1}{2} \sqrt{3 - 3 x^{2}}]

$cos^{-1}x+cos^{-1}\Bigg[\dfrac{x}{2}+\dfrac{1}{2}\sqrt{3-3x^2}\Bigg]$ is equal to

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$\dfrac{\pi}{3}$ for $x\in\Bigg[\dfrac{1}{2},1\Bigg]$
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$\dfrac{\pi}{3}$ for $x\in\Bigg[0,\dfrac{1}{2}\Bigg]$
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$2cos^{-1}x-cos^{-1}\dfrac{1}{2}$ for $x\in\Bigg[\dfrac{1}{2},1\Bigg]$
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$2cos^{-1}x-cos^{-1}\dfrac{1}{2}$ for $x\in\Bigg[0,\dfrac{1}{2}\Bigg]$

Explanation

Case 1:

0 \leq x \leq \frac{1}{2}

$0\leq x\leq\dfrac{1}{2}$ , then

c o s^{- 1} [\frac{x}{2} + \frac{1}{2} \sqrt{3 - 3 x^{2}}] = c o s^{- 1} [\frac{x}{2} + \frac{\sqrt{3}}{2} \sqrt{1 - x^{2}}] = c o s^{- 1} x - c o s^{- 1} \frac{1}{2}

$cos^{-1}\Bigg[\dfrac{x}{2}+\dfrac{1}{2}\sqrt{3-3x^2}\Bigg]=cos^{-1}\Bigg[\dfrac{x}{2}+\dfrac{\sqrt{3}}{2}\sqrt{1-x^2}\Bigg]=cos^{-1}x-cos^{-1}\dfrac{1}{2}$

Case 2:

\frac{1}{2} \leq x \leq 1

$\dfrac{1}{2}\leq x\leq1$ , then

c o s^{- 1} [\frac{x}{2} + \frac{1}{2} \sqrt{3 - 3 x^{2}}] = - c o s^{- 1} x + c o s^{- 1} \frac{1}{2}

$cos^{-1}\Bigg[\dfrac{x}{2}+\dfrac{1}{2}\sqrt{3-3x^2}\Bigg]=-cos^{-1}x+cos^{-1}\dfrac{1}{2}$

The domain of the function

\cos^{- 1} (2 x - 1)

$\cos ^{-1}(2x-1)$ is

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$[0,1]$
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$[-1,1]$
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$(-1,1)$
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$[0,\pi]$

Explanation

we have,

f (x) = \cos^{- 1} (2 x - 1)

$f(x)=\cos ^{-1}(2x-1)$

∵ - 1 \leq 2 x - 1 \leq 1

$\because -1 \le 2x -1 \le1$

\Rightarrow 0 \leq 2 x \leq 2

$\Rightarrow 0 \le 2x \le 2$

\Rightarrow 0 \leq x \leq 1

$\Rightarrow 0 \le x \le 1$

∴ x \in [0, 1]

$\therefore x \in [0,1]$

\cos (\sin^{- 1} \frac{2}{5} + \cos^{- 1} x) = 0

$\cos \left(\sin ^{-1}\dfrac{2}{5} + \cos ^{-1} x\right)=0$ then

x

$x$ is equal to

0%
$\dfrac{1}{5}$
0%
$\dfrac{2}{5}$
0%
$0$
0%
$1$

Explanation

we have

\cos (\sin^{- 1} \frac{2}{5} + \cos^{- 1} x) = 0

$\cos \left(\sin ^{-1}\dfrac{2}{5} + \cos ^{-1} x\right)=0$

\Rightarrow \sin^{- 1} \frac{2}{5} + \cos^{- 1} x = \cos^{- 1} 0

$\Rightarrow \sin ^{-1}\dfrac{2}{5}+ \cos ^{-1}x= \cos^{-1}0$

\Rightarrow \sin^{- 1} \frac{2}{5} + \cos^{- 1} x = \cos^{- 1} \cos \frac{π}{2}

$\Rightarrow \sin ^{-1}\dfrac{2}{5}+ \cos ^{-1}x= \cos^{-1} \cos \dfrac{\pi}{2}$

\Rightarrow \sin^{- 1} \frac{2}{5} + \cos^{- 1} x = \frac{π}{2}

$\Rightarrow \sin ^{-1}\dfrac{2}{5}+ \cos ^{-1}x= \dfrac{\pi}{2}$

\Rightarrow \cos^{- 1} x = \frac{π}{2} - \sin^{- 1} \frac{2}{5}

$\Rightarrow \cos ^{-1}x= \dfrac{\pi}{2} - \sin ^{-1}\dfrac{2}{5}$

\Rightarrow \cos^{- 1} x = \cos^{- 1} \frac{2}{5} [∵ \cos^{- 1} x + \sin^{- 1} x = \frac{π}{2}]

$\Rightarrow \cos ^{-1}x=\cos ^{-1} \dfrac{2}{5} \ \ [\because \cos^{-1}x+ \sin ^{-1}x=\dfrac{\pi}{2}]$

∴ x = \frac{2}{5}

$\therefore x=\dfrac{2}{5}$

The domain of trigonometric function can be restricted to any one of their branch (not necessarily principle value) in order to obtain their inverse functions.

0%
True
0%
False

The minimum value of

n

$n$ for which

\tan^{- 1} \frac{n}{π} > \frac{π}{4}, n \in N

$\tan^{-1}\dfrac {n}{\pi} > \dfrac {\pi}{4}, n\ \in \ N$ is valid is

5

$5$ .

0%
True
0%
False

Explanation

False

∵ \tan^{- 1} \frac{n}{π} > \frac{π}{4} \Rightarrow \frac{n}{π} > \tan \frac{π}{4}

$\because \tan^{-1} \dfrac {n}{\pi} > \dfrac {\pi}{4} \Rightarrow \dfrac {n}{\pi} > \tan \dfrac {\pi}{4}$

\Rightarrow \frac{n}{π} > 1 [∵ \tan \frac{π}{4} = 1]

$\Rightarrow \dfrac {n}{\pi} > 1 \left[\because \tan \dfrac {\pi}{4} =1\right]$

\Rightarrow n > π

$\Rightarrow n > \pi$
So, the minimum value of

n

$n$ is

4 [∵ n \in N a n d π = 3.14....]

$4 [\because n\in N\ and\ \pi=3.14....]$

All trigonometric functions have inverse over their respective domains.

0%
True
0%
False

The value of

\cot [c o s^{- 1} (\frac{7}{25})]

$\cot \left[cos^{-1}\left(\dfrac{7}{25}\right)\right]$ is

0%
$\dfrac{25}{24}$
0%
$\dfrac{25}{7}$
0%
$\dfrac {24} {25}$
0%
$\dfrac {7} {24}$

Explanation

we have

\cot [c o s^{- 1} (\frac{7}{25})]

$\cot \left[cos^{-1}\left(\dfrac{7}{25}\right)\right]$
Let

\cos^{- 1} \frac{7}{25} = x

$\cos ^{-1}\dfrac{7}{25}=x$

\Rightarrow \cos x = \frac{7}{25}

$\Rightarrow \cos x=\dfrac{7}{25}$
$$\therefore \sin x=\sqrt{1-\cos^2x=}\sqrt{1-\left(\dfrac{7}
{25}\right)^2}$$

= \sqrt{\frac{625 - 49}{625}} = \frac{24}{25}

$=\sqrt{\dfrac{625-49}{625}}=\dfrac{24}{25}$
$$\therefore \cot x=\dfrac{\cos x}{\sin x}= \dfrac{\dfrac{7}
{25}}{\dfrac{24}{25}}=\dfrac{7}{24}$$

\Rightarrow x = \cot^{- 1} (\frac{7}{24}) = \cot^{- 1} (\frac{7}{25})

$\Rightarrow x=\cot^{-1} \left(\dfrac{7}{24} \right)=\cot^{-1} \left(\dfrac{7}{25}\right)$

$$\therefore \cot \left(\cos^{-1}\dfrac{7}{25}\right)= \cot
\left(\cos^{-1}\dfrac{7}{24}\right)=\dfrac{7}{24} \ \ \left
[\because \cos^{-1}\dfrac{7}{24}=\cos^{-1}\dfrac{7}
{25}\right]$$
1656728_1795509_ans_4e362b3b1af34e6481c5b812a1d164f7.png

1656728_1795509_ans_4e362b3b1af34e6481c5b812a1d164f7.png

| x | \leq 1

$|x| \le 1$ , then

2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}})

$2 \tan ^{-1}x +\sin ^{-1} \left(\dfrac{2x}{1+x^2} \right)$ is equal to

0%
$4\ \tan^{-1}x$
0%
$0$
0%
$\dfrac{\pi}{2}$
0%
$\pi$

Explanation

We have,

| x | \leq 1

$|x| \le 1$ , then

2 \tan^{- 1} x + \sin^{- 1} (\frac{2 x}{1 + x^{2}})

$2 \tan ^{-1}x +\sin ^{-1} \left(\dfrac{2x}{1+x^2} \right)$

Let

x = \tan θ

$x=\tan \theta$

2 \tan^{- 1} \tan θ + \sin^{- 1} \frac{2 \tan θ}{1 + \tan^{2} θ}

$2 \tan^{-1} \tan \theta +\sin^{-1}\dfrac{2 \tan \theta}{1+ \tan^2 \theta}$

[∵ \tan^{- 1} (\tan x) = x]

$[\because \tan^{-1}(\tan x)=x]$

2 θ + \sin^{- 1} \sin 2 θ [∵ \sin 2 θ = \frac{2 \tan θ}{1 + \tan^{2} θ}]

$2 \theta +\sin^{-1} \sin 2 \theta \ \left[\because \sin 2 \theta =\dfrac{2 \tan \theta}{1+ \tan^2 \theta}\right]$

2 θ + 2 θ [∵ \sin^{- 1} (\sin x) = x]

$2 \theta +2 \theta \left[\because \sin^{-1} (\sin x)=x \right]$

4 θ [∵ θ \tan^{- 1}] 4 \tan^{- 1} x

$4 \theta \left[\because \theta \tan^{-1} \right]4 \tan^{-1} x$

The value of the expression

2 \sec^{- 1} 2 + \sin^{- 1} (\frac{1}{2})

$2 \sec ^{-1}2+ \sin^{-1}\left(\dfrac{1}{2}\right)$ is

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{5\pi}{6}$
0%
$\dfrac{7\pi}{6}$
0%
$1$

Explanation

We have

2 \sec^{- 1} 2 + \sin^{- 1} (\frac{1}{2})

$2 \sec ^{-1}2+ \sin^{-1}\left(\dfrac{1}{2}\right)$

2 \sec^{- 1} (\sec \frac{π}{3}) + \sin^{- 1} (\sin \frac{π}{6})

$2 \sec^{-1}(\sec \dfrac{\pi}{3})+\sin^{-1}(\sin \dfrac{\pi}{6})$

= 2. \frac{π}{3} + \frac{π}{6} [∵ \sec^{- 1} (\sec x) = x

$=2.\dfrac{\pi}{3}+\dfrac{\pi}{6} \ \ [\because \sec^{-1} (\sec x)=x$ and

\sin^{- 1} (\sin x) = x]

$\sin^{-1} (\sin x)=x]$

= \frac{4 π + π}{6} = \frac{5 π}{6}

$=\dfrac{4 \pi+\pi}{6}=\dfrac{5\pi}{6}$

The value of

\sin [2 \tan^{- 1} (.75)]

$\sin [2 \tan ^{-1} (.75)]$ is equal to

0%
$0.75$
0%
$1.5$
0%
$0.96$
0%
$\sin 1.5$

Explanation

we have,

\sin [2 \tan^{- 1} (.75)] = \sin (2 \tan^{- 1} \frac{3}{4}) [0.75 = \frac{75}{100} = \frac{3}{4}

$\sin [2 \tan ^{-1} (.75)] = \sin \left(2 \tan ^{-1}\dfrac{3}{4}\right) [0.75=\dfrac{75}{100}=\dfrac{3}{4}$

= \sin (\sin^{- 1} \frac{2. \frac{3}{4}}{1 + \frac{9}{16}}) = \sin [\sin^{- 1} \frac{3 / 2}{25 / 16}]

$=\sin \left(\sin ^{-1}\dfrac{2.\dfrac{3}{4}}{1+\dfrac{9}{16}}\right)=\sin \left[\sin ^{-1} \dfrac{3/2}{25/16}\right]$

= \sin [\sin^{- 1} (\frac{48}{50})] = \sin [\sin^{- 1} (\frac{48}{50})] = \frac{24}{25} = 0.96

$=\sin \left[\sin^{-1} \left(\dfrac{48}{50}\right)\right]=\sin \left[\sin^{-1} \left(\dfrac{48}{50}\right)\right]=\dfrac{24}{25}=0.96$

The value of

\cos^{- 1} (\cos \frac{3 π}{2})

$\cos ^{-1} \left(\cos \dfrac{3 \pi}{2}\right)$ is equal to

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{3\pi}{2}$
0%
$\dfrac{5\pi}{2}$
0%
$\dfrac{7\pi}{2}$

Explanation

We have,

\cos^{- 1} (\cos \frac{3 π}{2})

$\cos ^{-1} \left(\cos \dfrac{3 \pi}{2}\right)$

= \cos^{- 1} \cos (2 π \frac{π}{2}) [∵ \cos (2 π - \frac{π}{2})] = \cos \frac{π}{2}

$=\cos ^{-1}\cos \left( 2 \pi \dfrac{ \pi}{2}\right) \ \ \ \left[\because \cos \left(2 \pi -\dfrac{\pi}{2}\right)\right]= \cos \dfrac{\pi}{2}$

= \cos^{- 1} \cos (\frac{π}{2}) = \frac{π}{2} {∵ \cos^{- 1} (\cos x) = x, x \in [0, π]}

$=\cos ^{-1}\cos \left( \dfrac{ \pi}{2}\right) =\dfrac{ \pi}{2} \ \ \ \left\{\because \cos^{-1} (\cos x)=x, x \in [0,\pi] \right\}$
Note remember that,

\cos^{- 1} (\cos \frac{3 π}{2}) \neq \frac{3 π}{2}

$\cos ^{-1} \left(\cos \dfrac{3 \pi}{2}\right) \ne \dfrac{3 \pi}{2}$

∵ \frac{3 π}{2} \in (0, π)

$\because \dfrac{3 \pi }{2} \in (0, \pi)$

The value of the expression

(\cos^{- 1} x)^{2}

$(\cos^{-1}x)^2$ is equal to

\sec^{2} x

$\sec^2 x$ .

0%
True
0%
False

Explanation

False

∵ [\cos^{- 1} x]^{2} = {[\sec^{- 1} \frac{1}{x}]}^{2} \neq \sec^{2} x

$\because [\cos^{-1}x]^2 =\left[\sec^{-1}\dfrac 1x \right]^2\neq \sec^2 x$

\cos^{- 1} α + \cos^{- 1} β + \cos^{- 1} γ = 3 π

$\cos ^{-1} \alpha +\cos ^{-1} \beta +\cos ^{-1} \gamma = 3 \pi$ , then

α (β + γ) + β (γ + α) + γ (α + β)

$\alpha (\beta + \gamma) + \beta (\gamma + \alpha) + \gamma (\alpha + \beta)$

0%
$0$
0%
$1$
0%
$6$
0%
$12$

Explanation

we have,

\cos^{- 1} α + \cos^{- 1} β + \cos^{- 1} γ = 3 π

$\cos ^{-1} \alpha +\cos ^{-1} \beta +\cos ^{-1} \gamma = 3 \pi$

α (β + γ) + β (γ + α) + γ (α + β)

$\alpha (\beta + \gamma) + \beta (\gamma + \alpha) + \gamma (\alpha + \beta)$

we know that ,

0 \leq \cos^{- 1} x \leq π

$0 \le \cos^{-1} x \le \pi$
$$\Rightarrow \cos^{-1} \alpha +\cos^{-1} \beta +\cos^{-1}
\gamma = 3 \pi$$
If and only if, $$\cos^{-1}\alpha=\cos^{-1} \beta =\cos^{-1}
\gamma =\pi$$

\Rightarrow \cos π = α = β = γ

$\Rightarrow \cos \pi=\alpha=\beta=\gamma$

\Rightarrow - 1 = α = β = γ

$\Rightarrow -1=\alpha=\beta=\gamma$
$$\therefore \alpha (\beta + \gamma) + \beta (\gamma +
\alpha)+ \gamma (\alpha + \beta)$$

= - 1 (- 1 - 1) - 1 (1 - 1) - 1 (- 1 - 1)

$=-1(-1-1)-1(1-1)-1(-1-1)$

= 2 + 2 + 2 = 6

$=2+2+2=6$

The domain of the function defined by

f (x) = \sin^{- 1} x + \cos x

$f(x)=\sin^{-1}x+\cos x$ is

0%
$[-1, 1]$
0%
$[-1, \pi +1]$
0%
$( -\infty, \infty)$
0%
$\phi$

Explanation

(A) is the correct answer.

The domain of

\cos x

$\cos x$ is

R

$R$ and the domain of

\sin^{- 1}

$\sin^{-1}$ is

[- 1, 1]

$[-1, 1]$ .

∴

$\therefore$ The domain of

f (x) = \cos x + \sin^{- 1} x

$f(x)=\cos x+\sin^{-1}x$ is

R \cap [- 1, 1]

$R \cap [-1, 1]$ i.e.,

[- 1, 1]

$[-1, 1]$

\sin^{- 1} x + \sin^{- 1} y = \frac{π}{2}

$\sin^{-1}x+\sin^{-1}y=\dfrac{\pi}{2}$ , then the value of

\cos^{- 1} x + \cos^{- 1} y

$\cos^{-1}x+\cos^{-1}y$ is

0%
$\dfrac{\pi}{2}$
0%
$\pi$
0%
$0$
0%
$\dfrac{2\pi}{3}$

Explanation

(A) is the correct answer.

Given

\sin^{- 1} x + \sin^{- 1} y = \frac{π}{2}

$\sin^{-1}x+\sin^{-1}y=\dfrac{\pi}{2}$

We know that

\sin^{- 1} x + \cos^{- 1} y = \frac{π}{2}

$\sin^{-1}x+\cos^{-1}y=\dfrac{\pi}{2}$

∴ \cos^{- 1} x + \cos^{- 1} y = (\frac{π}{2} - \sin^{- 1} x) + (\frac{π}{2} - \sin^{- 1} y) = π - (\sin^{- 1} x + \sin^{- 1} y) = π - \frac{π}{2} = \frac{π}{2}

$\therefore\ \cos^{-1}x+\cos^{-1}y\\=\left( \dfrac {\pi}{2}-\sin^{-1}x\right) + \left( \dfrac {\pi}{2}-\sin^{-1}y \right) \\=\pi-(\sin^{-1}x+\sin^{-1}y)\\=\pi-\dfrac {\pi}{2}=\dfrac {\pi}{2}$

Hence,

\cos^{- 1} x + \cos^{- 1} y = \frac{π}{2}

$\cos^{-1}x+\cos^{-1}y=\dfrac {\pi}{2}$ .

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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