NCERT Exemplar Solutions Class 11 Maths Trigonometric Functions — Objective Type Q30–44

NCERT Exemplar Problems, Class 11 Mathematics — Trigonometric Functions, Objective Type questions 30 to 44. All 15 questions solved.

The first half of the Objective Type band. These questions mostly test whether you spot a shortcut: an AM-GM bound (30, 31), a complementary-angle product that collapses (34, 36), a standard compound-angle value (32, 38, 40), and a range or bound argument (33, 41). The rest are identity simplifications (35, 42 to 44). All printed answers agree with the working below.

Key insight. Most of these are one-line questions once the right fact is in hand: $\sin\theta+\mathrm{cosec}\,\theta$ has absolute value at least $2$; $\tan\theta\tan(90^\circ-\theta)=1$; $\cos90^\circ=0$ kills any product containing it; $a\cos x+b\sin x$ lies between $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$.

Question 30

Choose the correct answer. If $\sin\theta+\mathrm{cosec}\,\theta=2$, then $\sin^2\theta+\mathrm{cosec}^2\theta$ is equal to

Solution. Let $s=\sin\theta$. By AM-GM, $s+\frac1s\ge2$ for $s>0$, with equality only when $s=1$. Since the sum is exactly $2$, $\sin\theta=1$ and $\mathrm{cosec}\,\theta=1$, so $\sin^2\theta+\mathrm{cosec}^2\theta=1+1=2$. Another way: square the given equation, $s^2+\frac1{s^2}+2=4$, so $s^2+\frac1{s^2}=2$.

Question 31

Choose the correct answer. If $f(x)=\cos^2x+\sec^2x$, then

Solution. The two terms $\cos^2x$ and $\sec^2x=\frac1{\cos^2x}$ are positive and reciprocals of each other. By AM-GM, $\frac{\cos^2x+\sec^2x}2\ge\sqrt{\cos^2x\cdot\sec^2x}=1$, so $f(x)\ge2$, with equality when $\cos^2x=1$. Option (C) is an impossible range ($2<f<1$) and (A), (B) are false since $f(0)=2$.

Question 32

Choose the correct answer. If $\tan\theta=\dfrac12$ and $\tan\phi=\dfrac13$, then the value of $\theta+\phi$ is

Solution. Use the addition formula: $$\tan(\theta+\phi)=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=\frac{\frac56}{\frac56}=1.$$ Both $\tan\theta$ and $\tan\phi$ are positive and less than $1$, so each angle is acute and less than $45^\circ$; the sum is therefore in $(0,\frac\pi2)$, and $\tan(\theta+\phi)=1$ forces $\theta+\phi=\frac\pi4$ among the options.

Question 33

Choose the correct answer. Which of the following is not correct?

Solution. The range of each function decides this. $\sin\theta$ and $\cos\theta$ take any value in $[-1,1]$, so (A) and (B) are possible. $\tan\theta$ takes every real value, so (D) is possible. But $\sec\theta=\frac1{\cos\theta}$ always satisfies $|\sec\theta|\ge1$, so $\sec\theta=\frac12$ is impossible.

Question 34

Choose the correct answer. The value of $\tan1^\circ\tan2^\circ\tan3^\circ\ldots\tan89^\circ$ is

Solution. Pair the factor $\tan k^\circ$ with $\tan(90-k)^\circ$. Since $\tan(90^\circ-k^\circ)=\cot k^\circ$, each pair multiplies to $\tan k^\circ\cot k^\circ=1$. There are $44$ such pairs ($k=1,\dots,44$), and the middle factor $\tan45^\circ=1$. The product is $1$.

Question 35

Choose the correct answer. The value of $\dfrac{1-\tan^215^\circ}{1+\tan^215^\circ}$ is

Solution. This is the double-angle formula $\cos2A=\dfrac{1-\tan^2A}{1+\tan^2A}$ with $A=15^\circ$. So the value is $\cos30^\circ=\dfrac{\sqrt3}2$.

Question 36

Choose the correct answer. The value of $\cos1^\circ\cos2^\circ\cos3^\circ\ldots\cos179^\circ$ is

Solution. The product contains the factor $\cos90^\circ=0$, so the whole product is $0$. No other calculation is needed.

Question 37

Choose the correct answer. If $\tan\theta=3$ and $\theta$ lies in the third quadrant, then the value of $\sin\theta$ is

Solution. From $\tan\theta=3$, $\sec^2\theta=1+9=10$, so $|\cos\theta|=\frac1{\sqrt{10}}$ and $|\sin\theta|=|\tan\theta||\cos\theta|=\frac3{\sqrt{10}}$. In the third quadrant both $\sin\theta$ and $\cos\theta$ are negative, so $\sin\theta=-\dfrac3{\sqrt{10}}$.

Question 38

Choose the correct answer. The value of $\tan75^\circ-\cot75^\circ$ is equal to

Solution. Use $\tan75^\circ=\tan(45^\circ+30^\circ)=\dfrac{1+\frac1{\sqrt3}}{1-\frac1{\sqrt3}}=\dfrac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3$. Then $\cot75^\circ=\dfrac1{2+\sqrt3}=2-\sqrt3$. The difference is $(2+\sqrt3)-(2-\sqrt3)=2\sqrt3$.

Question 39

Choose the correct answer. Which of the following is correct?

Solution. In $\sin1^\circ$ the angle is one degree; in $\sin1$ the angle is one radian, which is about $57^\circ30’$ (the hint). Sine increases on $\left[0,\frac\pi2\right]$ and both $1^\circ$ and $57^\circ30’$ lie in that interval, so the larger angle has the larger sine: $\sin1^\circ<\sin1$. (Option (D) is also false: $\sin1^\circ\approx0.01745$ while $\frac\pi{180}\sin1\approx0.0147$, and the printed factor $\frac\pi{18^\circ}$ is dimensionally meaningless.)

Question 40

Choose the correct answer. If $\tan\alpha=\dfrac m{m+1}$ and $\tan\beta=\dfrac1{2m+1}$, then $\alpha+\beta$ is equal to

Solution. Compute $\tan(\alpha+\beta)$ and clear fractions: $$\tan(\alpha+\beta)=\frac{\frac m{m+1}+\frac1{2m+1}}{1-\frac m{m+1}\cdot\frac1{2m+1}}=\frac{m(2m+1)+(m+1)}{(m+1)(2m+1)-m}=\frac{2m^2+2m+1}{2m^2+2m+1}=1.$$ So $\alpha+\beta=\frac\pi4$ (for the values of $m$ where both angles are small and acute, the only option consistent with $\tan=1$ is $\frac\pi4$).

Question 41

Choose the correct answer. The minimum value of $3\cos x+4\sin x+8$ is

Solution. Write $3\cos x+4\sin x=5\sin(x+\varphi)$ with $\cos\varphi=\frac45$, $\sin\varphi=\frac35$, since $\sqrt{3^2+4^2}=5$. This ranges over $[-5,5]$, so the expression ranges over $[3,13]$, with minimum $8-5=3$.

Question 42

Choose the correct answer. The value of $\tan3A-\tan2A-\tan A$ is equal to

Solution. Start from $3A=2A+A$: $\tan3A=\dfrac{\tan2A+\tan A}{1-\tan2A\tan A}$. Clear the denominator: $\tan3A-\tan3A\tan2A\tan A=\tan2A+\tan A$. Rearranging, $$\tan3A-\tan2A-\tan A=\tan3A\tan2A\tan A.$$

Question 43

Choose the correct answer. The value of $\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$ is

Solution. The two angles $45^\circ+\theta$ and $45^\circ-\theta$ add to $90^\circ$, so they are complementary and $\cos(45^\circ-\theta)=\sin\left(90^\circ-(45^\circ-\theta)\right)=\sin(45^\circ+\theta)$. The difference is zero.

Question 44

Choose the correct answer. The value of $\cot\left(\dfrac\pi4+\theta\right)\cot\left(\dfrac\pi4-\theta\right)$ is

Solution. The angles $\frac\pi4+\theta$ and $\frac\pi4-\theta$ are complementary, so $\cot\left(\frac\pi4-\theta\right)=\tan\left(\frac\pi2-\left(\frac\pi4-\theta\right)\right)=\tan\left(\frac\pi4+\theta\right)$. The product is $\cot\left(\frac\pi4+\theta\right)\tan\left(\frac\pi4+\theta\right)=1$.

Common mistakes

• Question 30: solving $\sin\theta+\mathrm{cosec}\,\theta=2$ as a quadratic and getting lost. The AM-GM equality case ($\sin\theta=1$) gives the answer immediately.
• Question 31: choosing (B) or an upper bound. $\cos^2x\le1$ makes $\sec^2x\ge1$, but the sum is bounded below by $2$, not above by anything.
• Question 34: multiplying the sines and tangents numerically or thinking the product contains $\tan90^\circ$. The list stops at $89^\circ$, so every factor is finite.
• Question 37: giving the positive value $\frac3{\sqrt{10}}$ because $\tan\theta=3$ is positive. In the third quadrant $\tan$ is positive but $\sin$ and $\cos$ are both negative.
• Question 39: comparing $1^\circ$ with $1$ as if they were in the same unit. $1$ without a degree sign is one radian, nearly $57^\circ$.
• Question 42: picking (C), which looks like the pattern of $\tan(A+B+C)$ identities. The right product comes from $\tan(2A+A)$ only.

Practise next

• The same unit’s other exemplar pages: Short Answer Type, questions 1 to 19, Long Answer Type, questions 20 to 29, Objective Type, questions 45 to 59 and Fill in the Blanks, True or False and Matching, questions 60 to 76.
• NCERT textbook: Exercise 3.1 (radian and degree measure), Exercise 3.2 and Exercise 3.3.