MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 4

Given

$\sin A\, =\, \displaystyle \dfrac{3}{5}$ and

$\tan A$ is

$\displaystyle \frac{3}{m}$ , then

$m$ is:

0%
$1$
0%
$2$
0%
$4$
0%
$3$

Explanation

Given

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

$\sin A =\dfrac{P}{B} = \dfrac{3}{5}$
Thus,

$P = 3, H = 5$

By Pythagoras Theorem,

$H^2 = P^2 + B^2$

$5^2 = 3^2 + B^2$

$B= 4$ cm

Now,

$\tan A = \dfrac{P}{B} = \dfrac{3}{4}$

Find the smallest positive number p for which the equation

$cos (p sin x) = sin (p cos x)$ has a solution

$x \varepsilon [0, 2 \pi]$

0%
$\sqrt{2} \pi/4$
0%
$\sqrt{2} \pi/2$
0%
$\sqrt{2} \pi$
0%
$\pi/6$

Explanation

$\cos { \left( p\sin { x } \right) } =\sin { \left( p\cos { x } \right) }$

$\displaystyle \Rightarrow \cos { \left( p\sin { x } \right) } =\cos { \left( \frac { \pi }{ 2 } -p\cos { x } \right) }$

$\displaystyle \Rightarrow p\sin { x } =2n\pi \pm \left( \frac { \pi }{ 2 } -p\cos { x } \right)$

$\displaystyle \Rightarrow p\sin { x } +p\cos { x } =2n\pi +\frac { \pi }{ 2 }$ or

$\displaystyle p\sin { x } -p\cos { x } =2n\pi -\frac { \pi }{ 2 }$

$\displaystyle \Rightarrow p\sqrt { 2 } \left( \sin { \left( x+\frac { \pi }{ 4 } \right) } \right) =\frac { \left( 4n+1 \right) \pi }{ 2 }$
Or

$\displaystyle p\sqrt { 2 } \left( \sin { \left( x-\frac { \pi }{ 4 } \right) } \right) =\frac { \left( 4n-1 \right) \pi }{ 2 } \quad$

As

$\displaystyle -1\le \sin { \left( x+\frac { \pi }{ 4 } \right) } \le 1$

$\displaystyle \Rightarrow -p\sqrt { 2 } \le p\sqrt { 2 } \sin { \left( x+\frac { \pi }{ 4 } \right) } \le p\sqrt { 2 }$

$\displaystyle \Rightarrow -p\sqrt { 2 } \le \frac { \left( 4n+1 \right) \pi }{ 2 } \le p\sqrt { 2 }$ ...(1)

And as

$\displaystyle -1\le \sin { \left( x-\frac { \pi }{ 4 } \right) } \le 1$

$\displaystyle \Rightarrow -p\sqrt { 2 } \le p\sqrt { 2 } \sin { \left( x-\frac { \pi }{ 4 } \right) } \le p\sqrt { 2 }$

$\displaystyle \Rightarrow -p\sqrt { 2 } \le \frac { \left( 4n-1 \right) \pi }{ 2 } \le p\sqrt { 2 }$ ...(2)
(2) is always a subset of of first, therefore we have to consider only first
It is sufficient to consider

$n\ge 0$ , because for

$n>0$ , the solution will be same for

$n\ge 0$

If

$n\ge 0$

$\displaystyle -\sqrt { 2 } p\le \frac { \left( 4n+1 \right) \pi }{ 2 } \Rightarrow \frac { \left( 4n+1 \right) \pi }{ 2 } \le \sqrt { 2 } p$

$\displaystyle \sqrt { 2 } p\ge \frac { \pi }{ 2 } \Rightarrow p\ge \frac { \pi }{ 2\sqrt { 2 } } =\frac { \pi \sqrt { 2 } }{ 4 }$

Number of solutions of the equation

$tan x + sec x = 2 cos x$ lying in the interval

$[0, 2 \pi]$ is

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

Explanation

$tanx + secx = 2 cosx$

$\Rightarrow sinx + 1 = 2 cos^2x \Rightarrow sinx+1=2 - 2 sin^2x$

$\Rightarrow 2 sin^2x + sinx -1 = 0$

$\Rightarrow (2 sinx - 1) (sinx + 1) = 0$
but

$sinx = -1\Rightarrow x=\dfrac{3\pi}{2}$ ...........(1)

$sinx = \dfrac{1}{2} = sin\dfrac{\pi}{6}$
therefore general solution is,

$x = n\pi + (-1)^n.\dfrac{\pi}{6}$

$x = ......, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, ......$ ..........(2)
Therefore, number of solutions in the given interval are

$3$ .
Hence, option 'D' is correct.

The value of

$\sin^215^{\small\circ} + \sin^230^{\small\circ} + \sin^245^{\small\circ} + \sin^260^{\small\circ} + \sin^275^{\small\circ}$ is

0%
$1$
0%
$\displaystyle\frac{3}{2}$
0%
$\displaystyle\frac{5}{2}$
0%
$3$

Explanation

$(\sin^{2}75^{0}+\sin^{2}(15^{0}))+(\sin^{2}30^{0}+\sin^{2}60^{0})+\sin^{2}45^{0}$

$=(\cos^{2}15^{0}+\sin^{2}15^{0})+(\cos^{2}60^{0}+\sin^{2}60^{0})+\dfrac{1}{2}$

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

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

$=\dfrac{5}{2}$

Given that

$\tan (A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$ where

$A$ and

$B$ are acute angle.
Calculate

$A + B$ when

$\tan A = \dfrac{1}{2}, \tan B = \dfrac{1}{3}$ .

0%
$A+B =30^{\small\circ}$
0%
$A+B = 45^{\small\circ}$
0%
$A+B = 60^{\small\circ}$
0%
$A+B = 75^{\small\circ}$

Explanation

Given that

$\tan A = 1/2,\quad \tan B = 1/3$

$\tan(A+B) = \displaystyle\frac{\tan A + \tan B}{1-\tan A\tan B}$

$\quad = \displaystyle\frac{\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}}{1-\left(\displaystyle\frac{1}{2}\right)\left(\displaystyle\frac{1}{3}\right)} = \displaystyle\frac{5}{6}\times\displaystyle\frac{6}{5} = 1$

$\quad \therefore \tan(A+B) = 1 = \tan45^{\small\circ} \quad \Rightarrow A+B = 45^{\small\circ}$

$A = 60^{\small\circ}\ and\ B = 30^{\small\circ}$ , then verify each of the following:

$(i)\ \cos(A-B) = \cos A \cos B + \sin A \sin B$

$(ii)\cot(A+B) = \displaystyle\frac{\cot A\cot B - 1}{\cot A + \cot B}$

0%
$(i)\ True$
$(ii)\ False$
0%
$(i)\ False$
$(ii)\ False$
0%
$(i)\ True$
$(ii)\ True$
0%
$(i)\ False$
$(ii)\ True$

Explanation

$=cos(60^{0}-30^{0})$

$=cos(30^{0})$

$=\frac{\sqrt{3}}{2}$

$=LHS$
Now

$RHS$

$=cos60^{0}cos30^{0}+sin30^{0}sin60^{0}$

$=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}$

$=\frac{\sqrt{3}}{2}$

$=RHS$
Hence verified.

$=cot(60^{0}+30^{0})$

$=cot(90^{0})$

$=0$

$=LHS$
RHS

$=\frac{cot60^{0}.cot30^{0}-1}{cot60^{0}+cot30^{0}}$

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

$=0$

$Hence\ verified.$

Is LHS=RHS?

$\quad \sqrt{\displaystyle\frac{cosec\theta - 1}{cosec\theta + 1}}+\sqrt{\displaystyle\frac{cosec\theta + 1}{cosec\theta - 1}}=2\cos\theta$

Say true or false?

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

Simplifying the given expression, we get

$\dfrac{cosec\theta-1+cosec\theta+1}{\sqrt{cosec^{2}\theta-1}}$

$=\dfrac{2cosec\theta}{cot\theta}$

$=\dfrac{2cosec\theta}{cos\theta.cosec\theta}$

$=\dfrac{2}{cos\theta}$

$=2sec\theta$

Evaluate :

$\displaystyle\frac{2\tan30^{\small\circ}}{1-\tan^230^{\small\circ}}$

0%
$0$
0%
$1$
0%
$\sqrt2$
0%
$\sqrt3$

Explanation

We know that

$\tan 2A$

$=\dfrac{2\tan A}{1-\tan^{2}A}$
Here

$A=30^{0}$
Substituting, we get

$\dfrac{2\tan30^{0}}{1-\tan^{2}30^{0}}$

$=\tan(2(30^{0}))$

$=\tan(60^{0})$

$=\sqrt{3}$

Is LHS=RHS?

$\quad \displaystyle\frac{\cos\theta}{1+\sin\theta}+\displaystyle\frac{\cos\theta}{1-\sin\theta}=2cosec\theta$

Say true or false.

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

$\dfrac{cos(\theta)(1-sin\theta+1+sin\theta)}{1-sin^{2}\theta}$

$=\dfrac{cos(\theta)(2)}{cos^{2}\theta}$

$=\dfrac{2}{cos\theta}$

$=2sec\theta$

$\alpha + \beta = 90^{\small\circ}$ and

$\alpha = 2\beta$ , then

$\cos^2\alpha + \sin^2\beta$ equal

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

Explanation

It is given that

$\alpha=2\beta$
Hence

$\alpha+\beta=90^{0}$
Substituting, we get

$3\beta=90^{0}$
Hence

$\beta=30^{0}$
Therefore

$\alpha=60^{0}$
Hence

$\sin^{2}\beta+\cos^{2}\alpha$

$=\dfrac{1}{2}^{2}+\dfrac{1}{2}^{2}$

$=\dfrac{1}{4}+\dfrac{1}{4}$

$=\dfrac{1}{2}$

$p\sin x = q$ . If

$x$ is acute, then

$\displaystyle \sqrt{p^{2}-q^{2}}tan x$ is equal to

0%
$p$
0%
$q$
0%
$p q$
0%
$p + q$

Explanation

Given:

$p \sin x = q$
Now,

$\displaystyle \sqrt{p^{2}-q^{2}}\tan x$

$=\sqrt{p^{2}-p^{2}\sin ^{2}x.}\tan \:x$

$=\sqrt{p^{2}\left ( 1-\sin ^{2}x \right )}.\tan \:x$

$=p\sqrt{\left ( 1-\sin ^{2}x \right )}.\tan \:x\left .....( \because \cos ^{2}\theta +\sin ^{2}\theta =1 \right )$

$=P\sqrt{\cos ^{2}x}.\tan \:x$

$=p\cos \:x\tan \:x$

$=p\sin \:x$

$=q$

$\displaystyle \left | \tan x \right |=\tan x+\frac{1}{\cos x}(0\leq \times \leq 2\pi)$ has

0%
no solution
0%
one solution
0%
two solutions
0%
three solutions

Explanation

When,

$x \in [0, \frac{\pi}{2}]$ and

$x \in [\pi, \frac{3\pi}{2}]$
Thus,

$\tan x > 0$
hence,

$|\tan x | = \tan x$
hence,

$\displaystyle \left | \tan x \right |=\tan x+\frac{1}{\cos x}(0\leq \times \leq 2\pi)$
=

$\displaystyle \tan x =\tan x+\frac{1}{\cos x}$
=

$\cos x = 1$
=

$\cos x = \cos 90$
=

$x = 90^{\circ}$
But it is not possible since,

$\cos x \ne 0$ (

$\cos x$ being the denominator in the equation)

When,

$x \in [\frac{\pi}{2}, \pi ]$ and

$x \in [\frac{3\pi}{2}, 2\pi]$
Thus,

$\tan x < 0$
hence,

$|\tan x | = - \tan x$
hence,

$\displaystyle \left | \tan x \right |=\tan x+\frac{1}{\cos x}$
=

$2 \tan x = \frac{1}{\cos x}$
=

$\sin x = \frac{1}{2}$
This is not possible, as

$\sin x$ is not positive in these intervals of value of

$x$

Hence, the equation has no solution for any value of x.

$8 \tan A = 15$ , then the value of

$\displaystyle \frac{\sin A -\cos A}{\sin A+\cos A}$ is:

0%
$\displaystyle \frac{7}{23}$
0%
$\displaystyle \frac{11}{23}$
0%
$\displaystyle \frac{13}{23}$
0%
$\displaystyle \frac{17}{23}$

Explanation

$\displaystyle 8\tan A=15 \Rightarrow \tan A=\frac{15}{8}$

Now,

$\dfrac{\sin A-\cos A}{\sin A+\cos A}$
Multiply and divide by

$\cos A$

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

$=\dfrac{\tan A-1}{\tan A+1}$

Now, substitute the value of

$\displaystyle \tan A$

$= \dfrac{\dfrac{15}{8} -1}{\dfrac{15}{8}+1}$

$= \dfrac{7}{23}$

The number of solutions of the equation

$\displaystyle\frac { \sec { x } }{ 1 - \cos { x } } = \displaystyle\dfrac { 1 }{ 1 - \cos { x } }$ in

$\left[ 0, 2\pi \right]$ is equal to

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

Explanation

$\displaystyle\frac { \sec { x } }{ 1-\cos { x } } =\frac { 1 }{ 1-\cos { x } }$
For

$1-\cos { x } \neq 0\Rightarrow \cos { x } \neq 1$ ..(1)
Gives

$\sec { x } =1\Rightarrow \cos { x } =1$ ...(2)
From (1) and (2) we get no solution.

$x=2\sin ^{ 2 }{ \theta }$ ,

$y=2\cos ^{ 2 }{ \theta } +1$ , then the value of

$x+y$ is:

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

Explanation

Given,

$x=2{ \sin }^{ 2 }\theta ,y=2{ \cos }^{ 2 }\theta +1$

To find:

$(x+y)$

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

$=2({ \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta )+1$ .....(As

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

$=2+1$

$=3$

The solution set of

$(5+4\,cos\,\theta)\;(2\,cos\,\theta+1)=0$ in the interval

$[0,2\pi]$ is

0%
$\begin{Bmatrix}\displaystyle\frac{\pi}{3},\displaystyle\frac{2\pi}{3}\end{Bmatrix}$
0%
$\begin{Bmatrix}\displaystyle\frac{\pi}{3},{\pi}\end{Bmatrix}$
0%
$\begin{Bmatrix}\displaystyle\frac{2\pi}{3},\displaystyle\frac{4\pi}{3}\end{Bmatrix}$
0%
$\begin{Bmatrix}\displaystyle\frac{2\pi}{3},\displaystyle\frac{5\pi}{3}\end{Bmatrix}$

Explanation

$(5+4\cos\,\theta)\;(2\cos\,\theta+1)=0$

$\Rightarrow \cos\,\theta=-\displaystyle\frac{5}{4}$ (not possible)

$\cos\,\theta=-\displaystyle\frac{1}{2}$

$\Rightarrow \cos\theta=-\displaystyle\frac{1}{2}$

$\Rightarrow \cos \theta=-\cos {\dfrac{\pi}{3}}=\cos(\pi\pm\dfrac{\pi}{3})$

$\Rightarrow \theta=\displaystyle\frac{2\pi}{3},\displaystyle\frac{4\pi}{3}$ (II and III quad, cosine is negaitve)

Which of the following is / are the value (S) of the expression?
sin A(1+ tan A) + cos A (1+ cot A) ?
1. sec A + cosec A
2. 2 cosec A ( sin A + cos A )
3. tan A + cot A
Select the correct answer using the code given below.

0%
1 only
0%
1 and 2 only
0%
2 only
0%
1 and 3 only

Explanation

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

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

$(\sin A + \cos A) (\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})$
=

$(\sin A + \cos A) (\frac{\sin^2 A + \cos^2 A}{\sin A + \cos A}$
=

$\sec A + cosec A$
Thus, 1 satisfies the given equation.

$\displaystyle \sin x+\sin ^{2}x=1$ then the value of

$\displaystyle \cos ^{2}x+\cos ^{4}x$ is equal to

0%
1
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{3\sqrt{3}}$
0%
$\displaystyle \frac{3\sqrt{5}-5}{2}$

Explanation

Given,

$\sin x + \sin^2 x = 1$

$\Rightarrow \sin x = 1 - \sin^2 x$

$\Rightarrow \sin x = \cos^2x$

Hence,

$\cos^2 x + \cos^4x$

$=$

$\sin x + \sin^2 x$

$=$

$1$

....(Given,

$\sin x + \sin^2 x = 1$ )

$\displaystyle \cos \theta =\frac{5}{13}$ , where

$\theta$ being an acute angle, then the value of

$\dfrac{\cos \theta +5\cot \theta }{\text {cosec}\ \theta -\cos \theta }$ will be

0%
$\displaystyle \frac{169}{109}$
0%
$\displaystyle \frac{155}{109}$
0%
$\displaystyle \frac{385}{109}$
0%
$\displaystyle \frac{95}{109}$

Explanation

Given,

$\cos\theta=\dfrac{5}{13}$
Then

$\sin\theta=\dfrac{12}{13}$
Hence,

$\tan\theta=\dfrac{12}{5}$ and

$\text {cosec} \theta=\dfrac{13}{12}$
Substituting in the equation, we get

$=\dfrac{\frac{5}{13}+\frac{5.5}{12}}{\frac{13}{12}-\frac{5}{13}}$

$=\dfrac{5(12)+25(13)}{13(13)-5(12)}$

$=\dfrac{60+300+25}{169-60}$

$=\dfrac{385}{109}$

Without using trigonometric tables evaluate:-

$\displaystyle \frac{\cos ^{2}20^{\circ}+\cos ^{2}70^{\circ}}{\sec ^{2}50^{\circ}-\cot ^{2}40^{\circ}}+2\: cosec ^{2}58^{\circ}-2\cot 58^{\circ}\tan 32^{\circ}-4\tan 13^{\circ}\tan 37^{\circ}\tan 45^{\circ}\tan 53^{\circ}\tan 77^{\circ}$

0%
1
0%
2
0%
-1
0%
-2

Explanation

$\dfrac{\cos ^2 20^{\circ}+\cos ^2 70^{\circ}}{\sec ^2 50^{\circ}-\cot ^2 40^{\circ}}+2\csc ^2 58^{\circ}-2\cot 58^{\circ}\tan 32^{\circ}-4\tan 13^{\circ}\tan 37^{\circ}\tan 45^{\circ}\tan 53^{\circ}\tan 77^{\circ}$

$=\dfrac{\cos ^2 20^{\circ}+\sin ^2 20^{\circ}}{\sec ^2 50^{\circ}-\tan ^2 50^{\circ}}+2\csc ^2 58^{\circ}-2\cot 58^{\circ}\cot 58^{\circ}-4\tan 13^{\circ}\cot 13^{\circ} \cot 53^{\circ} \tan 53^{\circ} \tan 45^{\circ}$

$=\dfrac{1}{1}+2\csc ^2 58^{\circ}-2\cot 58^{\circ}\cot 58^{\circ}-4\times 1\times 1\times 1$

$=1+2(\csc ^2 58^{\circ}-\cot ^258^{\circ})-4$

$=1+2-4$

$=-1$

$\displaystyle \frac{\tan ^{2}\theta }{1+\sec \theta }+1$ equals to

0%
$\displaystyle \tan \theta$
0%
$\displaystyle \frac{1}{\cos \theta }$
0%
$\displaystyle \sec \theta -1$
0%
$\displaystyle \sec \theta +\tan \theta$

Explanation

$\displaystyle \frac{\tan ^{2}\theta }{1+\sec \theta }+1=\frac{\sec^2\theta-1 }{1+\sec \theta }+1$

$=\dfrac{(\sec\theta+1)(\sec\theta-1) }{1+\sec \theta }+1$

$=\sec\theta-1 +1$

$=\sec\theta$

$=\dfrac1{\cos\theta}$

Option B is correct.

$\displaystyle \sqrt{\frac{1-\sin \theta }{1+\sin \theta}}$ is equal to ............

0%
$\displaystyle cosec\ \theta -\cot \theta$
0%
$\displaystyle \tan \theta -\sec \theta$
0%
$\displaystyle\sec \theta -\tan \theta$
0%
$\displaystyle \cot \theta -cosec\ \theta$

Explanation

$\sqrt { \dfrac { 1-\sin { \theta } }{ 1+\sin { \theta } } }$

$=\sqrt { \dfrac { \left( 1-\sin { \theta } \right) \left( 1-\sin { \theta } \right) }{ \left( 1+\sin { \theta } \right) \left( 1-\sin { \theta } \right) } }$

$=\sqrt { \dfrac { { \left( 1-\sin { \theta } \right) }^{ 2 } }{ 1-\sin ^{ 2 }{ \theta } } }$

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

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

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

$=\sec { \theta } -\tan { \theta }$

Hence, the answer is

$\sec { \theta } -\tan { \theta }.$

The value of

$\displaystyle \frac{(1+\tan ^{2}\theta )}{(1+\cot ^{2}\theta )}$ is

0%
$\displaystyle \tan ^{2}\theta$
0%
$\displaystyle \cot ^{2}\theta$
0%
$\displaystyle \sec ^{2}\theta$
0%
$\displaystyle \text{cosec } ^{2}\theta$

Explanation

The value of

$\displaystyle \dfrac {(1+\tan ^{2}\theta )}{(1+\cot ^{2}\theta )}$ is

$=\displaystyle \dfrac {\left (1+\frac {\sin ^{2}\theta}{\cos ^{2}\theta} \right)}{\left (1+\frac {\cos ^{2}\theta}{\sin ^{2}\theta} \right)}$

$=\displaystyle \dfrac {\left (\frac {\cos ^{2}\theta + \sin ^{2}\theta}{\cos ^{2}\theta} \right)}{\left (\frac {\sin ^{2}\theta + \cos ^{2}\theta}{\sin ^{2}\theta}\right )}$

$=\displaystyle \dfrac {\left (\frac {1}{\cos ^{2}\theta}\right )}{\left (\frac {1}{\sin ^{2}\theta} \right)}$

$=\displaystyle \frac {\sin ^{2}\theta}{\cos ^{2}\theta}$

$=\displaystyle \tan^{2}\theta$
Hence, option A is correct.

The

$\triangle ABC$ has a right angle at C. If

$\displaystyle \sin A=\frac{2}{3}$ then

$\displaystyle \tan B$ is

0%
$\displaystyle \frac{3}{5}$
0%
$\displaystyle \frac{\sqrt{5}}{3}$
0%
$\displaystyle \frac{2}{\sqrt{5}}$
0%
$\displaystyle \frac{\sqrt{5}}{2}$

Explanation

$\Rightarrow \sin { A } =\dfrac { 2 }{ 3 } =\dfrac { BC }{ AB }$

In right angled

$\triangle ABC,$

$\Rightarrow AC^2+BC^2=AB^2$

$\Rightarrow AC^2=AB^2-BC^2$

$=9-4$

$\Rightarrow AC=\sqrt {5}$

$\Rightarrow \tan B=\dfrac{AC}{BC}=\dfrac{\sqrt {5}}{2}$

Hence, the answer is

$\dfrac{\sqrt {5}}{2}.$

The simplification of

$\displaystyle \sqrt{\frac{1+\cos A}{1-\cos A}}$ gives

0%
$\displaystyle \text{cosec } A+\cot A$
0%
$\displaystyle \text{cosec } A-\cot A$
0%
$\displaystyle \frac{1+\cos A}{\sin A}$
0%
Both A and C.

Explanation

$\displaystyle \sqrt{\dfrac{1+\cos A}{1-\cos A}}=\sqrt{\dfrac{1+\cos A}{1-\cos A}\times\dfrac{1+\cos A}{1+\cos A}}$

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

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

$=\displaystyle \dfrac{1+\cos A}{\sin A}$ ...(i)

$=\displaystyle \dfrac{1}{\sin A}+\dfrac{\cos A}{\sin A}$

$=\displaystyle \text{cosec} A+\cot A$ ...(ii)

From (i) and (ii) , Option D is correct.

The expression

$\displaystyle (1-\tan A+\sec A)(1-\cot A+\cos \sec A)$ has value

0%
-1
0%
0
0%
+1
0%
+2

Explanation

$\displaystyle \left(1-\tan A+\sec A\right)\left(1-\cot A+\text{cosec } A\right)$

$=\displaystyle \left(1-\frac {\sin A}{\cos A}+\frac {1}{cos A}\right)\left(1-\frac {\cos A}{\sin A}+\frac {1}{\sin A}\right)$

$=\displaystyle \left(\frac {\cos A- \sin A + 1}{\cos A}\right)\left(\frac {\sin A -\cos A + 1}{\sin A}\right)$

$=\displaystyle \left(\frac {\cos A- \sin A + 1}{\cos A}\right)\left(\frac {-\left(\cos A - \sin A -1\right)}{\sin A}\right)$

$=\displaystyle - \left(\frac {\left(\cos A- \sin A\right) + 1}{\cos A}\right)\left(\frac {\left(\cos A - \sin A\right) -1}{\sin A}\right)$

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

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

$=\displaystyle - \frac {\left(1- 2\sin A \cos A\right) - 1}{\sin A \cos A}$

$=\displaystyle - \frac {- 2\sin A \cos A }{\sin A \cos A} = 2$

Option D is correct.

In a right angled

$\displaystyle \Delta ABC$ right angled at

$B$ the ratio of

$AB$ to

$AC$ is

$\displaystyle 1:\sqrt{5}$ then

$\displaystyle 3\tan \theta +5\sec ^{2}\theta$ is

0%
$\displaystyle \frac{2}{\sqrt{5}}$
0%
$\displaystyle 3+\sqrt{5}$
0%
$\displaystyle \dfrac{25}4+\frac{3}{2}$
0%
$\displaystyle \frac{\sqrt{5+1}}{2}$

Explanation

$h=\sqrt { 5 } x$

$p=x$

$b=\sqrt { (\sqrt { 5 } )^ 2-x^ 2 } =\sqrt { 4{ x }^{ 2 } } =2x$

$\tan \theta =\dfrac { p }{ b } =\dfrac { 1x }{ 2x } =\dfrac { 1 }{ 2 }$

$\sec { \theta } =\dfrac { h }{ b } =\dfrac { \sqrt { 5 } x }{ 2x } =\dfrac { \sqrt { 5 } }{ 2 }$

$3\tan \theta +5\sec ^{ 2 }{ \theta =3\times } \dfrac { 1 }{ 2 } +5{ \times \left(\dfrac { \sqrt { 5 } }{ 2 } \right) }^{ 2 }=\dfrac { 3 }{ 2 } +\dfrac { 25 }{ 4 }$

The value of

$\displaystyle \frac{\sin ^{2}53+\cos ^{2}53}{\sec ^{2}37-\tan ^{2}37 }$ is

0%
$1$
0%
$2$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{3}{2}$

Explanation

Given,

$\displaystyle \frac{\sin ^{2}53+\cos ^{2}53}{\sec ^{2}37-\tan ^{2}37 }$

We know,

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

$\sec^2 \theta- \tan^2 \theta =1$

$\therefore \dfrac{\sin^2 53+ \cos^2 53}{\sec^2 37- \tan^2 37}=1$

Value of

$(1+\tan {\theta}+\sec {\theta})(1+\cot {\theta}-co\sec {\theta})$ is:

0%
$1$
0%
$-1$
0%
$2$
0%
$-4$

Explanation

$(1+\tan {\theta}+\sec {\theta})(1+\cot {\theta}-co\sec {\theta})$

$\left( 1+\cfrac { \sin { \theta } }{ \cos { \theta } } +\cfrac { 1 }{ \cos { \theta } } \right) \left( 1+\cfrac { \cos { \theta } }{ \sin { \theta } } -\cfrac { 1 }{ \sin { \theta } } \right)$

$\left( \cfrac { \cos { \theta } +\sin { \theta } +1 }{ \cos { \theta } } \right) \left( \cfrac { \sin { \theta } +\cos { \theta } -1 }{ \sin { \theta } } \right)$

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

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

$=\cfrac { 1+2\sin { \theta } \cos { \theta } -1 }{ \cos { \theta } \sin { \theta } } \quad =\cfrac { 2\sin { \theta \cos { \theta } } }{ \cos { \theta } \sin { \theta } } =2$

$\displaystyle \sec \theta =2,$ evaluating

$\displaystyle \frac{1-\tan \theta }{1+\tan \theta }$ gives

0%
$\displaystyle -\sqrt{3}$
0%
$\displaystyle \sqrt{3}+1$
0%
$\displaystyle2 -\sqrt{3}$
0%
$\displaystyle\frac{ \sqrt{3}+1}{2}$

Explanation

we know that,

$\sec^2\theta-\tan^2\theta=1$

$\Rightarrow \tan^2\theta=\sec^2\theta-1$

$\Rightarrow \tan^2\theta=4-1$

$[given\,\sec\theta=2]$

$\Rightarrow \tan\theta=\sqrt3$
Now,

$\displaystyle \dfrac{1-\tan \theta }{1+\tan \theta }=\dfrac{1-\sqrt3}{1+\sqrt3}\times\dfrac{1-\sqrt3}{1-\sqrt3}$

$\displaystyle=\frac{1+3-2\sqrt3}{1-3}$

$\displaystyle=\frac{4-2\sqrt3}{-2}$

$\displaystyle={\sqrt3-2}$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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