MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 2

The value of

$\operatorname { \sec } ^ { 2 } A \operatorname { \tan } ^ { 2 } B - \tan ^ { 2 } A \operatorname { \sec } ^ { 2 } B$ is

0%
$\operatorname { \tan } ^ { 2 } B - \tan ^ { 2 } A$
0%
$\operatorname { \sec } ^ { 2 } B + \operatorname { \sec } ^ { 2 } A$
0%
$\tan ^ { 2 } B - \operatorname { \sec } ^ { 2 } A$
0%
$\operatorname { \sec } ^ { 2 } B - \tan ^ { 2 } A$

Explanation

$\left( 1+{ \tan }^{ 2 }A \right) { \tan }^{ 2 }B-{ \tan }^{ 2 }A\left( 1+{ \tan }^{ 2 }B \right)$

$={ \tan }^{ 2 }B+{ \tan }^{ 2 }A{ \tan }^{ 2 }B-{ \tan }^{ 2 }A-{ \tan }^{ 2 }A{ \tan }^{ 2 }B$

$={ tan }^{ 2 }B-{ tan }^{ 2 }A\quad \left[ A \right]$

Which of the following is the principal value of

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

0%
$\dfrac { \pi }{ 3 }$
0%
$\dfrac { \pi }{ 6 }$
0%
$\dfrac { 2\pi }{ 3 }$
0%
$\dfrac { 3\pi }{ 3 }$

Explanation

we know

$cos^{-1}(-x) = \pi -cos^{-1}x.$

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

$= \pi -cos^{-1}(cos\dfrac{\pi }{3}).$

$= \pi -\dfrac{\pi }{3}.$

$= \dfrac{2\pi }{3}$

$\therefore$ principal value of

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

$\dfrac{2\pi }{3}.$

$cos x=b$ , then for what b do the roots of the equation form an AP ?

0%
$-1$
0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { \sqrt { 3 } }{ 2 }$
0%
None of these

Explanation

$\lim_{x\rightarrow 0} \frac{log(1+x^{3})}{sin^{3}x} = \lim_{x\rightarrow 0} \frac{log(1+x^{3})}{x^{3}} \frac{x^{3}}{sin^{3}x}$

$= \lim_{x\rightarrow 0} \frac{d}{dx} \frac{log(1+x^{3})}{\frac{d}{dx}}\times 1$

$= \lim_{x\rightarrow 0} \frac{\frac{1}{1+x^{3}}(3x)^{2}}{3x^{2}}\times 1$

$=1\times 1 = 1$

1210808_1354323_ans_c493e5015c164246aa4ff08b5f397bbb.jpg

The principle solutions of

$\sin \theta = - \frac{1}{2}$ are

0%
$\pi /4$
0%
$11\pi /6$
0%
$\pi /3$
0%
$\frac{{7\pi }}{6}$

Explanation

The principle solution is the solution of trignometric angle in [0,2

$\pi$ ]

So There are two principle solutions (

$\pi$ +

$\dfrac{\pi}{6}$ ) and (

$2\pi$ -

$\dfrac{\pi}{6}$ )=

$\dfrac{7\pi}{6}$ and

$\dfrac{11\pi}{6}$

$\sin A(1+\tan A) + \cos A(1+\cot A) = \sec A + \text{cosec} A.$

0%
True
0%
False

Explanation

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

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

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

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

$=\sec A + cosec \ A$

$\dfrac{\cos(90-A)\sin(90-A)}{\tan (90-A)}$ =

0%
$\sin^{2}A$
0%
$\cos^{2}A$
0%
$\sin A$
0%
$1$

Explanation

$= \dfrac{cos\,(90-A)\,sin\,(90-A)}{tan(90-A)}$

$= \dfrac{sin\,A \times cos\,A}{\dfrac{cos\,A}{sin\,A}} = \dfrac{sin^{2}A\times cos\,A}{cos\,A}$

$= sin^{2}A$

The principle solution of the

$Cos\theta = - \frac{1}{2}$ is

0%
$2\pi /3$
0%
$\pi /6$
0%
$4\pi /3$
0%
$\frac{{7\pi }}{6}$

Explanation

The values of

$\theta$ between 0 and

$2\pi$ at which

$Cos\theta=\dfrac{-1}{2}$ are

$\dfrac{2\pi}{3}$ and

$\dfrac{4\pi}{3}$

So Principal solutions are

$\dfrac{2\pi}{3}$ and

$\dfrac{4\pi}{3}$

$\sec^{2} A+ \csc^{2} A=\sec^{2} A \csc^{2} A$

0%
True
0%
False

Explanation

${sec}^2 A+\text{csc}^2 A=\dfrac{1}{\cos^2 A}+\dfrac{1}{\sin^2 A}$

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

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

$=\text{sec}^2 A\text{csc}^2 A$

What is the value of

$\cfrac{\tan{A}-\sin{A}}{\sin^3{A}}$ ?

0%
$\cfrac{\sec{A}}{1-\cos{A}}$
0%
$\cfrac{\sec{A}}{1+\cos^2{A}}$
0%
$\cfrac{\sec{A}}{1+\cos{A}}$
0%
None of these

Explanation

$\dfrac{\tan A-\sin A}{\sin^3 A}=\dfrac{\dfrac{\sin A}{\cos A}-\sin A}{\sin^3 A}=\dfrac{\dfrac{1-\cos A}{\cos A}}{\sin^2 A}=\dfrac{\text{sec} A(1-\cos A)}{1-\cos ^2 A}$

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

$=\dfrac{\text{sec}A}{1+\cos A}$

The principle solution of equation

$\cot x = - \sqrt 3$ is

0%
$\dfrac{\pi }{3}$
0%
$\dfrac{2\pi }{3}$
0%
$\dfrac{\pi }{6}$
0%
$\dfrac{5\pi }{6}$

Explanation

Given

$\cot x=-\sqrt{3}$

$\tan x=\dfrac{1}{\cot x}$

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

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

We know that

$\tan 30^o=1/\sqrt{3}$

We find the value of x where

$\tan$ is negative

$\tan$ is negative in

$2$ and

$4^{th}$ quadrant.

Value in

$2^{nd}$ Quadrant

$=180^o-30^o=150^o$

Value in

$4^{th}$ Quadrant

$=360^o-30^o=330^o$

So, principal solution are

$x=150^o$ and

$x=330^o$

$x=150 \times \pi/180$ and

$x=330\times \pi/180$

$x=5\pi/6$ and

$x=11\pi/6$

To find general solution

Let

$\tan x=\tan y$ ………..

$(1)$

$\tan x=-1/\sqrt{3}$ ………..

$(2)$

From

$(1)$ &

$(2)$

$\tan y=-1/\sqrt{3}$

$\tan y=\tan 5\pi/6$

$y=5\pi/6$

Since

$\tan x=\tan y$

General solution is

$x=n\pi +y$ where

$n\in z$ .

Hence

$x=n\pi +5\pi /6$ where

$n\in z$ .

The range of the function

$f(x)=\left (\cos ^2x+4\sec ^2x\right )$ is

0%
$[4,\infty )$
0%
$[0,\infty )$
0%
$[5,\infty )$
0%
$(0,\infty )$

Explanation

As we know that

$AM\ge GM$

$\implies \dfrac{\cos^2 x+\text{sec}^2 x}{2}\ge \sqrt{(\cos^2 x)(4\text{sec}^2 x)}$

$\implies \dfrac{f(x)}{2}\ge \sqrt{4}$

$\implies f(x)\ge 4$

So the range of

$f(x)$ is

$[4,\infty)$

$\displaystyle 1^c =?$

0%
$\displaystyle 56^027'22''$
0%
$\displaystyle 57^016'22''$
0%
$\displaystyle 55^018'32''$
0%
$\displaystyle 57^026'32''$

Explanation

$\displaystyle \pi^c = 180^0 \\ \Rightarrow 1^c = \left(\dfrac{180}{\pi}\right)^0 = \left(180 \times \dfrac{7}{22}\right)^0 = \left(\dfrac{630}{11}\right)^0 = 57^016'22''$ (app.).

If tan

$\theta =\dfrac{1}{\sqrt{5}}$ and

$\theta$ lies in the first quadrant, the value of

$\cos\theta$ is

0%
$\dfrac{1}{\sqrt{6}}$
0%
$\dfrac{\sqrt{5}}{\sqrt{6}}$
0%
$\dfrac{-1}{\sqrt{6}}$
0%
$\dfrac{-\sqrt{5}}{\sqrt{6}}$

Explanation

Given

$\theta$ lies in first quadrant

$\implies \cos \theta>0$

$\tan \theta=\dfrac{1}{\sqrt{5}}$

As we know that

$\sec^2 \theta=1+\tan^2 \theta=1+\bigg(\dfrac{1}{\sqrt{5}}\bigg)^2=1+\dfrac{1}{5}=\dfrac{6}{5}$

$\implies \sec \theta=\dfrac{\sqrt{6}}{\sqrt{5}}$

$\cos \theta=\dfrac{1}{\sec \theta}=\dfrac{1}{\dfrac{\sqrt{6}}{\sqrt{5}}}=\dfrac{\sqrt{5}}{\sqrt{6}}$

$0^0 < x < 45^0,$ then consider the following statements :

Assertion (A):

$\dfrac{1}{1+sin^{2} x}-\dfrac{1}{1+sec^{2} x}\neq \dfrac{1}{1+cos^{2} x}-\dfrac{1}{1+cosec^{2}x}$

Reason (R) :

$sin x \neq cos x$

of these statements :

0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true, but R is not the correct explanation of A.
0%
A is true, but R is false.
0%
A is false, but R is true.

Explanation

Consider the first expression

$\dfrac{1}{1+sin^2x}-\frac{1}{1+sec^2x}$

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

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

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

$=\dfrac{1-sin^2xcos^2x}{2+sin^2xcos^2x}$ ...(i)
Let us now consider second expression

$\dfrac{1}{1+cos^2x}-\frac{sin^2x}{1+sin^2x}$

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

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

$=\dfrac{1-cos^2xsin^2x}{2+sin^2xcos^2x}$

$= (ii)$
However

$sinx=cosx\:at\:x=45^0$

$sinx$ is not equal to

$cosx$ for

$0<x<45^0$
Hence reason is true assertion is wrong

$1)$ Principal value of

$\cos\theta=-1$ is

$\pi$

$2)$ Principal value of

$\sin\theta=0$ is

$\pi$
Which of the above statement is correct?

0%
Only I
0%
Only II
0%
Both I and II
0%
Neither I or II

Explanation

we know that ,

$\cos\theta=-1\Rightarrow \theta=\cos^{-1}(-1 ) \Rightarrow\theta=\pi$

as Range of Principal Value of

$\cos^{-1}$ is

$[0,\pi]$

But,

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

as Range of Principal Value of

$\sin^{-1}$ is

$[\dfrac{-\pi}{2},\dfrac{\pi}{2}]$

Therefore, Only,

$(1)$ is Correct

Let

$n$ be a positive integer such that

$\displaystyle \sin(\frac{\pi}{2^{n}})+\cos(\frac{\pi}{2^{n}})=\frac{\sqrt{n}}{2}$

0%
$<4$
0%
$n>8$
0%
$4\leq n<8$
0%
$6\leq n\leq 8$

Explanation

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

$1+2\sin \cfrac{\pi}{2^n}\cos\cfrac{\pi}{2^n}=\dfrac{n}{4}$ ...... $[\cos^{2}\theta +\sin^{2}\theta =1]$

$\displaystyle \Rightarrow 1+\sin \cfrac{\pi }{2^{n-1}}=\cfrac{n}{4}$

$\displaystyle \sin(\cfrac{\pi }{2^{n-1}})=\cfrac{n-4}{4}$
AS $n=+ve. \neq 1$ and $\sin\theta\le 1$
$\displaystyle 0 < \cfrac{n-4}{4} \leq 1$
$\therefore 4 < n \leq 8$

$\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}$ are the principal values of following trigonometric equations
1)

$\sin\theta=-\dfrac 1{\sqrt{2}}$
2)

$\displaystyle \cos\theta=-\frac{\sqrt{3}}{2}$
3)

$\tan\theta=\sqrt{3}-2$

0%
$p_{1}< p_{2}< p_{3}$
0%
$p_{1}< p_{3}< p_{2}$
0%
$p_{3}< p_{1}< p_{2}$
0%
$p_{2}< p_{3}< p_{1}$

Explanation

$\sin\theta=\dfrac{-1}{\sqrt2}\Rightarrow \theta=\sin^{-1}(\dfrac{-1}{\sqrt2})\Rightarrow \theta=\dfrac{-\pi}{4}$
as Range of Principal Value of

$\sin^{-1}$ is

$[\dfrac{-\pi}{2},\dfrac{\pi}{2}]$

$\cos\theta=-\dfrac{\sqrt3}{2}\Rightarrow \theta=\cos^{-1}(-\dfrac{\sqrt3}{2} ) \Rightarrow\theta=\dfrac{2\pi}{3}$
as Range of Principal Value of

$cos^{-1}$ is

$[0,\pi]$

$\tan\theta=\sqrt3-2\Rightarrow \theta=\tan^{-1}(\sqrt3-2 ) \Rightarrow\theta=\dfrac{\pi}{2}+\dfrac{\pi}{12}=\dfrac{7\pi}{12}$
as Range of Principal Value of

$\tan^{-1}$ is

$(\dfrac{-\pi}{2},\dfrac{\pi}{2})$

Since,

$\dfrac{-\pi}{4}<\dfrac{7\pi}{12}<\dfrac{2\pi}{3}$

Therefore,

$p_1<p_3<p_2$

The number of solutions of the equation $8{\tan ^2}\theta + 9 = 6\sec \theta$ in the interval $(\frac{-\pi}{2}, \frac{\pi}{2})$

0%
Two
0%
Four
0%
Zero
0%
None of these

Explanation

${(\tan x)}^2$ =

${(\sec x)}^2$ -1

So, 8 [

${(\sec x)}^2$ -1]+9=6

${\sec x}$

${(\sec x)}^2$ -8+9=6

${\sec x}$

${\sec x}$ =8

${(\sec x)}^2$ +1

${(\sec x)}^2$ -6

${\sec x}$ +1=0

So, we get a quadratic in

$"\sec x".$

on solving we get, there does not exist any value of 'x' for which the aove equation is satisfied.

Therefore, zero solution.

(sec⁡

$\alpha$ and

$\beta$ are two different solutions lying between

$\displaystyle \frac{-\pi}{2}$ and

$\displaystyle \frac{\pi}{2}$ of the equation

$2t\mathrm{a}n\theta+\mathrm{S}\mathrm{e}\mathrm{c}\theta=2$ then

$t\mathrm{a}n\alpha+t\mathrm{a}n\beta$

0%
0
0%
1
0%
4/3
0%
8/3

Explanation

The given equation is

$2\tan{\theta}+\sec{\theta}=2$

$\Rightarrow \sec{\theta}=2-2\tan{\theta}$

Squaring on both sides, we get

${\sec}^{2}{\theta}=4{\left(1-\tan{\theta}\right)}^{2}$

${\sec}^{2}{\theta}=4\left(1+{\tan}^{2}{\theta}-2\tan{\theta}\right)$

$1+{\tan}^{2}{\theta}=4\left(1+{\tan}^{2}{\theta}\right)-8\tan{\theta}$

$3{\tan}^{2}{\theta}-8\tan{\theta}+3=0$

Since

$\alpha$ and

$\beta$ are the roots of the equation,

$\therefore \tan{\alpha}+\tan{\beta}=\dfrac{8}{3}$

For all values of

$\theta,\cos\theta.\text{cosec }\theta\sqrt{\sec^2\theta-1}$ is

0%
$1$
0%
$\cos^2\theta$
0%
$\cot\theta$
0%
$\tan\theta$

Explanation

${ \cos }\theta\text{cosec }\theta \sqrt { { \sec }^{ 2 }\theta -1 }$

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

$=\cot\theta .\tan\theta \\ =1$ .

$sec \theta - tan \theta=k,$ then the value of

$sec \theta + tan \theta$ is

0%
$(1 - k)$
0%
$(1+k)$
0%
$\dfrac{1}{k}$
0%
$1-\dfrac{1}{k}$

Explanation

Given,

$sec \theta - tan \theta =k................(i)$
Also,

$sec^2 \theta - tan^2 \theta =1.............(ii)$
Dividing equation (ii) by (i), we get

$\dfrac{sec^{2} \theta-tan^{2} \theta}{sec \theta-tan \theta}=\dfrac{1}{k}$
or

$sec \theta + tan \theta=\dfrac{1}{k}$

Which of the following is/are FALSE

$\forall \theta \in R$ ?

0%
$\sin \theta =1 - \cos \theta$
0%
$\sec \theta - \tan \theta =\dfrac{1}{\sec \theta + \tan \theta}$
0%
$\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta$
0%
$\sin \dfrac{\pi}{3}=\cos\dfrac{\pi}{6}$

Explanation

Dividing LHS by RHS we get,

$A\rightarrow \dfrac{sin\theta}{1-cos\theta}\neq 1$

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

$C\rightarrow cosec^2\theta -cot^2\theta=1$

$D\rightarrow \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{\sqrt3}{2}}=1$

Therefore, Answer is

$sin\theta=1-cos\theta$

$\tan\theta=\dfrac{3}{4}$ and

$0< \theta, < 90^0,$ then the value of

$\sin\theta \cos \theta$ is

0%
$\dfrac{1}{5}$
0%
$\dfrac{9}{5}$
0%
$\dfrac{12}{25}$
0%
$\dfrac{25}{12}$

Explanation

Given,

$\tan \theta=\dfrac{3}{4}$

$\therefore \sec \theta=\sqrt{1+\tan^2\theta}$

$=\sqrt{1+\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}$

$\cos \theta=\dfrac{4}{5}$

$\therefore \sin \theta \cos \theta = \tan \theta \cos^2 \theta$

$=\dfrac{3}{4}\times \dfrac{16}{25}=\dfrac{12}{25}$

$\dfrac{3-4 \sin^2 \theta}{\cos^2 \theta}$ is equal to

0%
$3 - \cot^2 \theta$
0%
$3 + \cot^2 \theta$
0%
$3 - \tan^2 \theta$
0%
$3+ \tan^2 \theta$

Explanation

The given expression is

$\dfrac { 3-4\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =\dfrac { 3\times 1-4\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =\dfrac { 3\times (\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } )-4\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =\dfrac { 3\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =3-\dfrac { \sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \\ =3-\tan ^{ 2 }{ \theta } .$

The value of

$(\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta)^2$ is

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

Explanation

$(\cos\theta+\sin\theta)^2+(\cos\theta-\sin\theta)^2\\ =\cos^2\theta+\sin^2\theta+2\sin\theta \cos\theta+\cos^2\theta+\sin^2\theta-2\sin\theta \cos\theta$

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

$=2(1)=2$

$\dfrac{\tan^{2}\theta}{(1+\sec \theta)^{2}}$ is equal to

0%
$\left ( \dfrac{1-\cos \theta}{1+\cos \theta} \right )$
0%
$\left ( \dfrac{1+\cos \theta}{1-\cos \theta} \right )$
0%
$\left ( \dfrac{\cos \theta-1}{\cos \theta+1} \right )$
0%
$\left ( \dfrac{\cos \theta+1}{\cos \theta-1} \right )$

Explanation

Given,

$\dfrac { \tan ^{ 2 }{ \theta } }{ { \left( 1+\sec ^{ 2 }{ \theta } \right) }^{ 2 } } =\dfrac { \sec ^{ 2 }{ \theta } -1 }{ { \left( 1+\sec ^{ 2 }{ \theta } \right) }^{ 2 } }$

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

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

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

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

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

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

Explanation

$\sec ^{ 2 }{ \theta +\text{cosec} ^{ 2 }{ \theta } } \\ =\dfrac { 1 }{ \cos ^{ 2 }{ \theta } } +\dfrac { 1 }{ \sin ^{ 2 }{ \theta } } =\dfrac { \sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta } } }{ \sin ^{ 2 }{ \theta .\cos ^{ 2 }{ \theta } } } \\ =\dfrac { 1 }{ \sin ^{ 2 }{ \theta .\cos ^{ 2 }{ \theta } } } =\text{cosec} ^{ 2 }{ \theta .\sec ^{ 2 }{ \theta } }$

The expression

$\sin^6 \theta+\sin^4 \theta \cos^2 \theta - \sin^2 \theta \cos^4 \theta - \cos^6 \theta$ can be reduced to

0%
$\sin^2 \theta + \cos^2 \theta$
0%
$\sin^3 \theta - \cos^3 \theta$
0%
$\sin^4 \theta + \cos^4 \theta$
0%
$\sin^2 \theta -\cos^2 \theta$

Explanation

${ \sin }^{ 6 }\theta +{ \sin }^{ 4 }\theta { \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta { \cos }^{ 4 }\theta -{ \cos }^{ 6 }\theta$

$={ \sin }^{ 4 }\theta ({ \sin }^{ 2 }\theta +{\cos }^{ 2 }\theta )-{ \cos }^{ 4 }\theta ({ \sin }^{ 2 }\theta+{\cos }^{ 2 }\theta )$

$={ \sin }^{ 4 }\theta -{ \cos }^{ 4 }\theta \left[ \because { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1 \right]$

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

$={ \sin }^{ 2 }\theta -{ \cos }^{ 2 }\theta$

Hence, option $D$ is the correct answer.

Which of the following statements is not correct?

0%
$cos^4 \theta sin^4 \theta - cos^2 \theta - sin^2 \theta$
0%
$1+tan^2 \theta = sec^2 \theta$
0%
$sin 40^0 + cos50^0 = 2 sin 40^0$
0%
$sin^2 \theta + cos^2 \theta=2$

Explanation

From the options , it is observed that option D is not correct.

Range of

$\sin\theta$ is

$-1 \le \sin\theta \le 1$ for

$\theta \in R$

$-1 \le \cos\theta \le 1$ for

$\theta \in R$

or,

$\sin^{2}\theta \le 1$ and

$\cos^{2}\theta \le 1$

But

$\sin\theta$ and

$\cos\theta$ both are constant , it is clearly observed from the graph.

So,

$\sin\theta$ and

$\cos\theta$ both cannot be 1 simultaneously.

It is also an identity that,

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

The value of

$\cos^4\theta-\sin^4 \theta+2 \sin^2 \theta$ is equal to

0%
$0$
0%
$\sin^2 \theta$
0%
$\cos^2\theta$
0%
$1$

Explanation

$\cos^4\theta-\sin^4\theta+2\sin^2\theta$

$=(\cos^2\theta)^2-\sin^4\theta +2\sin^2\theta$

$=(1 - \sin^2\theta)^2 - \sin^4\theta +2\sin^2\theta$

$=1 - 2\sin^2\theta + \sin^4\theta-\sin^4\theta +2\sin^2\theta$

$=1$
Hence answer is D

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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