If $$\sin\theta=\alpha$$ then, $$\tan\theta$$=......... .

$$\dfrac{\alpha}{1-\alpha^{2}}$$

MCQ Questions for Class 10 Maths Introduction To Trigonometry Quiz 11

$\sin \theta = \dfrac {3}{5}$ , then the value of

$\text{cosec}\, \theta$ is

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

Explanation

Given

$\sin \ \theta=\dfrac{3}{5}$

We know that

$\text{cosec}\, \theta=\dfrac{1}{\sin \ \theta}=\dfrac{1}{\dfrac{3}{5}}$

$\therefore$

$\text{cosec}\, \theta=\dfrac{5}{3}$

The value of tan

$1^o$ tan

$2^o$ tan

$3^o$ .... tan

$89^o$ is.

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

We have to find the value of

$tan 1^{0} tan 2^{0} tan 3^{0} .....tan 89^{0}$

We know that

$tan 89^{0}=tan(90^{0}-1^{0})=cot 1^{0}$

Similarly

$tan 88^{0}=cot 2^{0}$ and so on

Finally we have

$tan46^{0}=cot44^{0}$

So we get

$tan1^{0} tan 2^{0}........tan 89^{0}=(tan1^{0}cot1^{0})(tan2^{0}cot2^{0}).........(tan44^{0}cot44^{0})(tan45^{0})$

$=(1)(1).....(1)(1)=1$

$sin^2 45 - sin^2 60 = x cos^2 45$ then x = ..............

0%
3/2
0%
3/4
0%
-1/2
0%
2

Explanation

$sin^2 45 - sin^2 60 = x cos^2 45$

$\Rightarrow \left( \dfrac{1}{\sqrt 2} \right )^2 - \left( \dfrac{\sqrt 3}{2} \right )^2 = x \left( \dfrac{1}{\sqrt 2} \right )^2$

$\Rightarrow \left( \dfrac{\sqrt 2}{2} \right )^2 - \left( \dfrac{\sqrt 3}{2} \right )^2 = x \left( \dfrac{\sqrt 2}{2} \right )^2$

$\Rightarrow \dfrac{2}{4} - \dfrac{3}{4} = \dfrac{2}{4} x$

$\Rightarrow - \dfrac{1}{4} = \dfrac{1}{2} x$

$\Rightarrow x = - \dfrac{1}{2}$

$sin^2 (3x + 45) + cos^2 (2x + 60) = 1$ then

$x =$ ...............

0%
$60$
0%
$30$
0%
$15$
0%
$0$

Explanation

$sin^2 (3x + 45)+ cos^2 (2x + 60) = 1$

$\therefore sin^2 (3x + 45) = 1 - cos^2 (2x + 60)$

$\therefore sin^2 (3x + 45) = sin^2 (2x + 60)$ ............. [since

$1 - cos^2 \theta = sin^2 \theta$ ]

$\therefore 3x + 45 = 2x + 60$

$\therefore x = 15$

Which of the following pair is a correct trignometric inter-relationship?

(1) $\cos { \theta }$	(a) $\cfrac { 1 }{ \tan { \theta } }$
(2) $\tan { \theta }$	(b) $\cfrac { 1 }{ co\sec { \theta } }$
(3) $\cot { \theta }$	(c) $\cfrac { 1 }{ \sec { \theta } }$
(4) $\sin { \theta }$	(d) $\cfrac { 1 }{ \cot { \theta } }$

0%
$1-d,2-e,3-b,4-a$
0%
$1-b,2-a,3-e,4-d$
0%
$1-c,2-d,3-a,4-b$
0%
$1-e,2-b,3-c,4-d$

Explanation

We can write

$\cos \theta$ as

$\dfrac{1}{\dfrac{1}{\cos \theta}}=\dfrac{1}{\sec \theta}$

$\therefore \cos \theta=\dfrac{1}{\sec \theta}$ ..........

$(1)$

Now,

$\tan \theta=\dfrac{\sin \theta}{\cos \theta}=\dfrac{1}{\dfrac{\cos \theta}{\sin \theta}}=\dfrac{1}{\cot \theta}$

$\therefore \tan \theta=\dfrac{1}{\cot \theta}$ ........

$(2)$

$\cot{\theta} =\cfrac{1}{\tan{\theta}}$ ............

$(3)$

$\sin { \theta } =\dfrac{1}{\dfrac{1}{\sin \theta}}=\dfrac{1}{\text{cosec}{\theta}}$ ......

$(4)$

Hence, the correct answer is

$1-c,2-d,3-a,4-b$

The value of

$\sin 60^{\circ} \cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ}$ is equal to

0%
$\sec 90^{\circ}$
0%
$\tan 90^{\circ}$
0%
$\cos 60^{\circ}$
0%
$\sin 90^{\circ}$

Explanation

$\sin { 60° } \cos { 30° } +\cos { 60° } \sin { 30° }$

$\Rightarrow \dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 } +\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } =\dfrac { 3 }{ 4 } +\dfrac { 1 }{ 4 } =\dfrac { 4 }{ 4 } =1$

Also,

$\sin { 90° } =1$

$\therefore \sin { 60° } \cos { 30° } +\cos { 60° } \sin { 30° }=\sin { 90° }$

Hence, the answer is

$\sin { 90° }.$

$\sin { 7\theta } =\cos { 2\theta }$ for acute angles

$7\theta$ and

$2\theta$ , then

$\theta =$

0%
$10$
0%
$90$
0%
$20$
0%
$30$

Explanation

Given :

$\sin { 7\theta } =\cos { 2\theta }$

$\implies \sin { 7\theta } =\sin { (90^o-2\theta ) }$

$\implies 7\theta =90^o-2\theta$

$\implies 9\theta =90^o$

$\implies \theta =10^o$

Hence, the answer is

$10^o$ .

The value of

$\tan 26^{\circ}\tan 64^{\circ}$ is

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

Explanation

$\tan { 26° } \tan { 64° } =\tan { \left( 90°-64° \right) } \tan { 64° }$

$=\cot { 64° } \tan { 64° }$

$(\because \tan (90-x) =\cot x)$

$=1.$

$(\because\tan x\cdot \cot x=1 )$

Hence,

$\tan { 26° } \tan { 64° } =1.$

$7\cos ^{ 2 }{ \theta } +3\sin ^{ 2 }{ \theta } =4$ , then

$\cot { \theta } =$

0%
$\cfrac { 7 }{ 3 }$
0%
$7$
0%
$\sqrt { 3 }$
0%
$\cfrac { 1 }{ \sqrt { 3 } }$

Explanation

$7\cos^{2}{\theta}+3\sin^{2}{\theta} =4$

$\implies 7\cos^{2}{\theta}+3\sin^{2}{\theta}=4\left(\cos^{2}{\theta}+\sin^{2}{\theta}\right)$

$\implies 7\cos^{2}{\theta}+3\sin^{2}{\theta}=\left(4\cos^{2}{\theta}+4\sin^{2}{\theta}\right)$

$\implies 3\cos^{2}{\theta}=\sin^{2}{\theta}$

$\implies \tan^{2}{\theta} =3$

$\implies \tan{\theta}=\sqrt{3}$

$\implies \dfrac{1}{\tan \theta}=\dfrac{1}{\sqrt{3}}$
then

$\cot { \theta } =\cfrac { 1 }{ \sqrt { 3 } }$

Hence, option D is correct.

$cos\, 25^0 +sin \, 25^0=k$ , then

$cos\, 20^0$ is equal to

0%
$\dfrac{k}{\sqrt{2}}$
0%
$-\dfrac{k}{\sqrt{2}}$
0%
$\pm \dfrac{k}{\sqrt{2}}$
0%
none of these

Explanation

$\cos 25^{\circ}+\sin 25^{\circ}=k$

$\dfrac{1}{\sqrt{2}}\cos 25^{\circ}+\dfrac{1}{\sqrt{2}}\sin 25^{\circ}=\dfrac{k}{\sqrt{2}}$

$\sin (45^{\circ}+25^{\circ})=\dfrac{k}{\sqrt{2}}$

$\sin 70^{\circ}=\dfrac{k}{\sqrt{2}}$

$\sin (90^{\circ}-20^{\circ})=\dfrac{k}{\sqrt{2}}$

$\cos 20^{\circ}=\dfrac{k}{\sqrt{2}}$

The value of

$\sec^2 \theta + \text{cosec}^2 \theta$ is equal to :

0%
$\tan^2 \theta + \cot^2 \theta$
0%
$\sec^2 \theta. \text{cosec}^2 \theta$
0%
$\sec \theta .\text{cosec} \theta$
0%
$\sin^2 \theta. \cos^2 \theta$
0%
1

Explanation

The value of

$\sec^2 \theta + \text{cosec}^2 \theta$ is

$= \dfrac {1}{\cos^2 \theta} + \dfrac {1}{\sin^2 \theta}$

$= \dfrac { \sin^2 \theta + \cos^2 \theta} {\sin^2 \theta \cos^2 \theta}$

$= \dfrac {1}{ \sin^2 \theta \cos^2 \theta }$ .....[Since

$\sin^2\theta+\cos ^2\theta=1$ ]

$= \sec^2 \theta .\text{cosec}^2 \theta$

$\tan \theta + \cot \theta = 2$ , then the value of

$\tan^{2}\theta + \cot^{2}\theta$ is ____________.

0%
$4$
0%
$2$
0%
$\dfrac {3}{2}$
0%
$5$

Explanation

We know

$tan\theta=\dfrac{1}{cot\theta}$ ,

So,

$(tan\theta+cot\theta)^{2}=tan^{2}\theta+cot^{2}\theta+2$

$\Rightarrow (2)^{2}=tan^{2}\theta+cot^{2}\theta+2$

$\Rightarrow tan^{2}\theta+cot^{2}\theta=2^{2}-2$

$\Rightarrow tan^{2}\theta+cot^{2}\theta=2$

So, Option (B)

Evaluate:

$\displaystyle\frac{-\tan\theta \cot(90^o-\theta)+\sec\theta cosec(90^o-\theta)+\sin^235^o+\sin^255^o}{\tan 10^o\tan 20^o\tan 30^o \tan 70^o \tan 80^o}$ .

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

Explanation

as we know

$cot(90^0$

$-$

$\theta$

$)$

$=$

$tan$

$\theta$ and

$cosec(90^0$

$-$

$\theta$

$)$

$=$

$sec$

$\theta$

Numerator

$sec ^2$

$\theta$

$-$

$tan ^ 2$

$\theta$

$=1$ and

$sin 55^0 =$

$cos 35^0$

So using the above mentioned identity the numerator outcome is

$2$ and the denominator

$tan 80^0 =$

$cot 10^0$ and

$tan 10^0 * cot 10^0 = 1$ similarly for

$tan 20^0 cot 20^0 = 1$

so denominator is

$tan30^0 = 1 /$

$\sqrt{3}$

Hence answer is

$2*$

$\sqrt{3}$

Hence option B is correct.

$4\tan \theta = 3$ , then

$\dfrac {5\sin \theta + 3\cos \theta}{5\sin \theta - 3\cos \theta} =$

0%
$9$
0%
$5$
0%
$10$
0%
$4$

Explanation

$4\tan \theta = 3$

$\tan \theta = \dfrac {3}{4}$ . So,

$\sin \theta/cos \theta = \dfrac {3}{4}$

Hence,

$\dfrac {5\left (\dfrac {sin \theta}{cos \theta}\right ) + 3}{5\left (\dfrac {sin \theta}{cos \theta}\right ) - 3}$ .

$\dfrac {5\left (\dfrac {3}{4}\right ) + 3}{5\left (\dfrac {3}{4}\right ) - 3}$ .

$=9$

The value of

$\dfrac { \sin { A } }{ 1+\cos { A } } +\dfrac { \sin { A } }{ 1-\cos { A } }$ is

$({0}^{o}<A<{90}^{o})$ :

0%
$2\csc A$
0%
$2\sec A$
0%
$2\sin A$
0%
$2\cos A$

Explanation

$\dfrac { \sin { A } }{ 1+\cos { A } } +\dfrac { \sin { A } }{ 1-\cos { A } }$

$=\dfrac { \sin A (1-\cos A) +\sin A(1+\cos A)}{(1+\cos A)(1-\cos A)}$

$=\dfrac { \sin A-\sin A\cos A +\sin A+\sin A\cos A }{1-\cos^{2}A}$

$=\dfrac { 2\sin A }{\sin^{2} A}$

$=\dfrac { 2}{\sin A}$

$=2\csc A$

Hence the correct option is (A).

$\tan x = x - \dfrac{1}{4x}$ , then

$\sec x - \tan x$ is equal to:

0%
$2x$
0%
$-2x$
0%
$4x$
0%
$-4x$

Explanation

$\tan x = x - \dfrac{1}{4x}$

We know that,

$sec^2x=1+tan^2x$

$secx=\sqrt{1+tan^2x}$

So,

$sec^2x=1+(x-\dfrac{1}{4x})^2$

$=1+x^2-2.\dfrac1{4x}.x+\dfrac{1}{16x^2}$

$=x^2+2.\dfrac1{4x}.x+\dfrac{1}{16x^2}$

$=(x+\dfrac1{4x})^2$

$secx=\pm(x+\dfrac{1}{4x})$

So either,

$\sec x-\tan x=x+\dfrac{1}{4x}-( x - \dfrac{1}{4x})$

$=\dfrac{1}{2x}$

or,

$\sec x-\tan x$

$=-x-\dfrac{1}{4x}-( x - \dfrac{1}{4x})$

$=-2x$

If :

$\tan A=\dfrac {\sqrt {3}-1}{\sqrt {3}+1}$ , then :

$\cos A=$

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

Explanation

given

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

we know

$\sec ^{ 2 }{ A } =1+\tan ^{ 2 }{ A } \\ \Rightarrow \sec ^{ 2 }{ A } =1+\dfrac { 2-\sqrt { 3 } }{ 2+\sqrt { 3 } } =\dfrac { 4 }{ 2+\sqrt { 3 } } \\ \Rightarrow \cos ^{ 2 }{ A } =\dfrac { 2+\sqrt { 3 } }{ 4 }$

therefore

$\cos { A } =\dfrac { \sqrt { 2+\sqrt { 3 } } }{ \sqrt { 4 } } =\dfrac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } }$

option B

$(\sin A+\cos A)(1-\sin A \cos A)=\sin^3A+\cos^3 A$

0%
True
0%
False

What is the value of

$\dfrac{\sin A}{1- \cos A}-\cot A = \csc A$

0%
$1$
0%
$2$
0%
$5$
0%
None

Explanation

$\dfrac{{\sin A}}{{1 - \cos A}} - \cot A$

$= \dfrac{{\sin A}}{{1 - \cos A}} - \dfrac{{\cos A}}{{\sin A}}$

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

$= \dfrac{{\left( {1 - \cos A} \right)}}{{\left( {1 - \cos A} \right)\sin A}}$ ........

$[sin²θ+cos²θ=1]$

$= cosecA$ ..........

$\because cosec{A}=\dfrac{1}{\sin{A}}$

State true or false.

$\dfrac{\cos\theta}{1-\sin\theta}= \dfrac{\tan\theta+ \sec\theta-1}{\tan\theta+ \sec\theta-1}$

0%
True
0%
False

$\cos {\theta _1} = 2\cos {\theta _2},$ then

$\tan \dfrac{{{\theta _1} - {\theta _2}}}{2}$ .

$\tan \dfrac{{{\theta _1} + {\theta _2}}}{2}$ is equal to

0%
$\dfrac{1}{3}$
0%
$\dfrac{-1}{3}$
0%
$1$
0%
$-1$

Explanation

$\sin \dfrac{A-B}{2}\times \sin \dfrac{A+B}{2}=\sin^2 \dfrac{A}{2}-\sin^2\dfrac{B}{2}=1-\cos A-1+\cos B=\cos B-\cos A$

$\cos \dfrac{A-B}{2}\times \cos \dfrac{A+B}{2}=\cos^2 \dfrac{B}{2}-\sin^2\dfrac{A}{2}=1+\cos B-1+\cos A=\cos B+\cos A$

$\tan \dfrac{\theta_1-\theta_2}{2}\tan \dfrac{\theta_1+\theta_2}{2}=\dfrac{\sin \dfrac{\theta_1-\theta_2}{2}\sin \dfrac{\theta_1+\theta_2}{2}}{\cos\dfrac{\theta_1-\theta_2}{2}\cos \dfrac{\theta_1+\theta_2}{2}}=\dfrac{\cos \theta_2-\cos \theta_1}{\cos \theta_2+\cos \theta_1}=\dfrac{\cos \theta_2-2\cos \theta_2}{\cos \theta_2+2\cos \theta_2}=\dfrac{-\cos \theta_2}{3\cos \theta_2}=\dfrac{-1}{3}$

In a right angled triangle, the square of the hypotenuse is twice the product of the other two sides. Then the angle of the triangle is:

0%
$15^\circ$
0%
$30^\circ$
0%
$45^\circ$
0%
$60^\circ$

Explanation

$b^{2}sec^{2}\theta = b^{2}tan \theta$

$\dfrac{1}{cos^{2}\theta }= \dfrac{sin\theta }{cos\theta }$

$sin \theta = cos\theta$

$\theta = 45^{\circ}$

1063588_1102816_ans_6bc99738ed1845f7897125d35f21f352.jpeg

$\cos ^{ 2 }{ \theta } \left( 1+\tan ^{ 2 }{ \theta } \right) +\sin ^{ 2 }{ \theta } \left( 1+\cot ^{ 2 }{ \theta } \right) =2$

0%
True
0%
False

Explanation

$\cos^2 \theta(1+\tan^2 \theta)+\sin^2 \theta(1+\cot^2 \theta)=\cos^2 \theta(\text{sec}^2 \theta)+\sin^2 \theta(\text{cosec}^2 \theta)$

$=\cos^2 \theta\times \dfrac{1}{\cos^2 \theta}+\sin^2 \theta\times \dfrac{1}{\sin^2 \theta}$

$=1+1$

$=2$

So the given relation is

$\text{True}$

$\sin\theta=\alpha$ then,

$\tan\theta$ =......... .

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

Explanation

According to question,

$\sin\theta=\alpha$ ,

then,

$\cos\theta=\sqrt{1-\alpha^2}$

$\implies\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

$\implies\tan\theta=\dfrac{\alpha}{\sqrt{1-\alpha^2}}$

Solve

$\sin A(1+\tan A)+\cos A(1+\cot A)$ .

0%
$\sec A+cosec A$
0%
$\sin A+cos A$
0%
$\tan A+\cot A$
0%
$\cos A+\sin A$

Explanation

$sinA(1+tanA)+cosA(1+cotA)$

$\Rightarrow sinA\left( { 1+\dfrac { { \sin A } }{ { \cos A } } } \right) +cosA\left( { 1+\dfrac { { \cos A } }{ { \sin A } } } \right)$

$\Rightarrow \dfrac { { \sin A } }{ { \cos A } } \left( { \sin A+\cos A } \right) +\dfrac { { \cos A } }{ { \sin A } } \left( { \sin A+\cos A } \right)$

$\Rightarrow \left( { \sin A+\cos A } \right) \left[ { \dfrac { { { { \sin }^{ 2 } }A+{ { \cos }^{ 2 } }A } }{ { \sin A\cdot \cos A } } } \right] \\$

$\Rightarrow \dfrac { { \sin A } }{ { \sin A\cdot \cos A } } +\dfrac { { \cos A } }{ { \sin A\cdot \cos A } } \\$

$\Rightarrow \sec A+ cosec \, A \\$

$\tan A+\cot A=4,\ then\ {\tan}^{4}\ A+{\cot}^{4}\ A$ is equal to

0%
$110$
0%
$191$
0%
$80$
0%
$194$

Explanation

$\textbf{Expanding the given expression using simple trigonometric ratios}$

$\text{We have given,}$

$\displaystyle{\tan A+ \cot A=4}$

$\text{On squaring both sides we get,}$

$(\tan A+ \cot A)^{2}= 4^{2}$

$\Rightarrow \tan^{2}A+\cot^{2}A+ 2 (\tan A) (\cot A)= 16$

$\boldsymbol{[\because\tan A \cot A=1]}$

$\Rightarrow \tan^{2}A+ \cot^{2}A+2=16$

$\Rightarrow \tan^{2}A+ \cot^{2}A= 16-2$

$\Rightarrow \tan^{2}A+\cot^{2}A=14$

$\textbf{On squaring both sides we get,}$

$\Rightarrow (\tan^{2}A+\cot^{2}A)^{2}= (14)^{2}$

$\Rightarrow\tan^{4}A+ \cot^{4}A+2(\tan^{2}A)(\cot^{2}A)= 196$

$\boldsymbol{[\because\tan^2 A \cot^2 A=1]}$

$\Rightarrow\tan^{2}A+ \cot^{4}A= 196-2$

$= 194$

$\textbf{Hence option 'D' is correct}$

Check whether

$\dfrac {\cos 30^{o}+\sin 60^{o}}{1+\cos 60^{o}+\sin 30^{o}}=\dfrac {\sqrt {3}}{4}$ is true/false ?

0%
True
0%
False

Explanation

As we know that

$\cos 30^{\circ}=\dfrac{\sqrt{3}}{2},\sin 60^{\circ}=\dfrac{\sqrt{3}}{2},\sin 30^{\circ}=\cos 60^{\circ}=\dfrac{1}{2}$

$\dfrac{\cos 30^{\circ}+\sin 60^{\circ}}{1+\cos 60^{\circ}+\sin 30^{\circ}}=\dfrac{\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}}{1+\dfrac{1}{2}+\dfrac{1}{2}}=\dfrac{\sqrt{3}}{2}$

So the given relation is

$\text{False}$

$\sin^{4}\theta+2\sin^{2}\theta\left(1-\dfrac {1}{cosec^{2}\theta}\right)+ \cos^{4}\theta=$

0%
$1$
0%
$0$
0%
$1/2$
0%
$-1$

Explanation

$\displaystyle \rightarrow sin^{4}\theta +2sin^{2}\theta (1-\frac{1}{cosec^{2}\theta })+cos^{4}\theta$

$\displaystyle = sin^{4}\theta +2sin^{2}\theta (\frac{cosec^{2}\theta -1}{cosec^{2}\theta })+cos^{4}\theta$

$\displaystyle = sin^{4}\theta +2sin^{4}\theta \times \frac{cos^{2}\theta }{sin^{2}\theta }+cos^{4}\theta ...(cot^{2}\theta )$

$\displaystyle = sin^{4}\theta +2sin^{2}\theta cos^{2}\theta +cos^{4}\theta$

$\displaystyle = (sin^{2}\theta +cos^{2}\theta )^{2}$

$\displaystyle = 1^{2} = 1$

1121235_1165092_ans_05bf3bc2cdca40fa80f347794c51d5f7.jpeg

$sin60 ^0=\cfrac{tan30^0}{1+tan^{2}30^0}$ .

0%
True
0%
False

Explanation

As we know that

$\sin 2 A=\dfrac{2\tan A}{1+\tan^2 A}$

Put

$A=30^{\circ}$

$\implies \sin 2(30^{\circ})=\dfrac{2\tan 30^{\circ}}{1+\tan^2 30^{\circ}}$

$\implies \sin 60^{\circ}=\dfrac{2\tan 30^{\circ}}{1+\tan^2 30^{\circ}}$

So the Given relation is

$\text{False}$

$2 \tan 45^{\circ} + \cos 45^{\circ} - \sin 45 ^{\circ} =?$

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

We have,

$\sin45^\circ=\dfrac{1}{\sqrt2}$

$\cos45^\circ=\dfrac{1}{\sqrt2}$

$\tan45^\circ=1$

Substitute these values in the given expression to find its value,

$\begin{aligned}{}2\tan {45^\circ } + \cos {45^\circ } - \sin {45^\circ }&= 2 \times 1 + \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}\\ &= 2\end{aligned}$

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 11 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 11 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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