CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 11 - MCQExams.com

The value of$$tan^{2}(sec^{-1}3)+cot^{2}(cosec^{-1}4)$$ is -
  • 9
  • 16
  • 25
  • 23
$${ cos }^{ -1 }\left[ 2cot^{ -1 }\left( \surd 2-1 \right)  \right] $$ is equal to
  • $$\sqrt { 2 } -1$$
  • $$1-\sqrt { 2 } $$
  • $$\pi /4$$
  • $$3\pi /4$$
If $${ \cot }^{ -1 }(\sqrt { \cos a } )-{ \tan }^{ -1 }(\sqrt { \cos a } )=x,$$ then $$\sin x$$ is equal to
  • $${ \cot }\dfrac { a }{ 2 } $$
  • $${ \cot }^{ 2 }\dfrac { a }{ 2 } $$
  • $${\tan }^{ 2 }\dfrac { a }{ 2 } \quad $$
  • $$\tan a$$
$${ tan }^{ -1 }\left( \dfrac { \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1+{ -x }^{ 2 } }  }{ \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1{ -x }^{ 2 } }  }  \right) ,[x]\le \dfrac { 1 }{ \sqrt { 2 }  } ,is\quad equal\quad to$$
  • $$\dfrac { 1 }{ 2 } { cos }^{ -1 }{ x }^{ 2 }$$
  • $$\frac { \pi }{ 4 } -\dfrac { 1 }{ 2 } { cos }^{ -1 }{ x }^{ 2 }$$
  • $$\frac { \pi }{ 4 } +\dfrac { 1 }{ 2 } { cos }^{ -1 }{ x }^{ 2 }$$
  • $$\frac { \pi }{ 2 } -\dfrac { 1 }{ 2 } { cos }^{ -1 }{ x }^{ 2 }$$
If $$sin^{-1}x+sin^{-1}y=\cfrac {2 \pi}3$$, then cos^{-1}x+cos^{-1}y$$ equal to
  • $$\cfrac {2 \pi} 3$$
  • $$\cfrac {\pi}3$$
  • $$\cfrac {\pi}6$$
  • $$\pi$$
$$tan^{-1}(1-x^{2}-\frac{1}{x^{2}})+sin^{-1}(x^{2}+\frac{1}{x^{2}}-1)$$ is equal to 
  • $$\frac{\pi }{2}$$
  • $$\frac{\pi }{4}$$
  • $$\frac{3\pi }{4}$$
  • $$\pi $$
If $$cot^{1}\frac{1}{x}+cot^{-1}\frac{1}{y}+cot^{-1 }\frac{1}{z}=\frac{\pi }{2}$$ then.....
  • x+y+z=xyz
  • xy+yz+zx=1
  • x+y+z=1
  • $$x^{2}+y^{2}+z^{2}=1$$
If$${\cos ^{ - 1}}\left( {2{x^2} - 1} \right) = 2\pi  - 2{\cos ^{ - 1}}x,$$ then
  • $$x \in \left[ { - 1,0} \right]$$
  • $$x \in \left[ {0,1} \right]$$
  • $$x \in \left[ {0,\frac{1}{{\sqrt 2 }}} \right]$$
  • $$x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$$
$$4 sin^{-1}x+cos^{-1} x=\pi  $$ then x=
  • $$\frac{-1}{4}$$
  • $$\frac{1}{4}$$
  • $$\frac{-1}{2}$$
  • $$\frac{1}{2}$$
Evaluate : $$\sin \left(\dfrac {1}{2}\cos^{-1}\dfrac {4}{5}\right)$$ 
  • $$\dfrac {1}{\sqrt 5}$$
  • $$\dfrac {2}{\sqrt 5}$$
  • $$\dfrac {1}{\sqrt {10}}$$
  • $$\dfrac {2}{\sqrt {10}}$$
If $$\alpha =2\tan^{-1}(\sqrt{3-2\sqrt{2}})+\sin^{-1}\left(\dfrac{1}{\sqrt{6}-\sqrt{2}}\right), \beta =\cot^{-1}(\sqrt{3}-2)+\dfrac{1}{8}\sec^{-1}(-2)$$ & $$\gamma =\tan^{-1}\dfrac{1}{\sqrt{2}}+\cos^{-1}\dfrac{1}{\sqrt{3}}$$, then?
  • $$\alpha =\beta$$
  • $$\alpha +\beta =3\gamma$$
  • $$4(\beta -\gamma)=\alpha$$
  • $$\beta =\gamma$$
If $$\cos^{-1} x > \sin^{-1} x$$, then find the range of x

  • $$-1 \leq x < \dfrac {1}{\sqrt {2}}$$


  • $$0\leq < \dfrac {1}{\sqrt {2}}$$
  • $$\dfrac {1}{\sqrt {2}} < x \leq 1$$
  • $$x > 0$$
If $$\cot^{-1}\left(\dfrac {-1}{5}\right)=x$$ then $$\sin x=$$ ?
  • $$\dfrac {1}{\sqrt {26}}$$
  • $$\dfrac {5}{\sqrt {26}}$$
  • $$\dfrac {1}{\sqrt {24}}$$
  • $$none\ of\ these$$
The value of $$\tan^{-1}\Bigg(\dfrac{x\cos\theta}{1-x\sin\theta}\Bigg)-\cot^{-1}\Bigg(\dfrac{\cos\theta}{x-\sin\theta}\Bigg)$$  is
  • $$2\theta$$
  • $$\theta$$
  • $$\dfrac{\theta}{2}$$
  • independent of $$\theta$$
$$ tan^{-1} \left (\dfrac{c_1\,x - y}{c_1\,y + x} \right ) + tan^{-1} \left ( \dfrac{c_2 - c_1}{1 + c_2c_1} \right ) + tan^{-1} \left (\dfrac{c_3 - c_2}{1 + c_3c_2}  \right ) + $$ ..... $$ + tan^{-1}\left (\dfrac{1}{c_n}  \right ) = $$
  • $$ tan^{-1} \,(y/x) $$
  • $$ tan^{-1}\dfrac{x}{y} $$
  • $$ -tan^{-1}\left ( \dfrac{x}{y} \right ) $$
  • none of these
Let  $$\begin{vmatrix}tan^{-1}x & tan^{-1}2x & tan^{-1}3x\\ tan^{-1}3x & tan^{-1}x & tan^{-1}2x\\
tan^{-1}2x & tan^{-1}3x & tan^{-1}x \end{vmatrix}$$= 0, then the number of values of x satisfying the equation is 
  • $$1$$
  • $$2$$
  • $$3$$
  • $$4$$
Choose the correct answer :
$$ \tan^{-1}\sqrt{3} - \cot^{-1} (-\sqrt{3})  $$ is equal to  
  • $$ \pi $$
  • $$ - \dfrac{\pi }{2}$$
  • $$ 0 $$
  • $$ 2 \sqrt{3} $$
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