NCERT Exemplar Solutions Class 11 Maths Trigonometric Functions — Objective Type Q45–59

NCERT Exemplar Problems, Class 11 Mathematics — Trigonometric Functions, Objective Type questions 45 to 59. All 15 questions solved.

The second half of the Objective Type band. It leans on exact values of $18^\circ$, $36^\circ$, $54^\circ$ and $72^\circ$ (46, 48, 56), on sum-to-product (46, 49, 54), on the half-angle and double-angle formulae (52, 57, 58), and on a counting question (53) where one root must be rejected. All printed answers agree with the working below.

Key insight. Two exact values carry most of this page: $\sin18^\circ=\frac{\sqrt5-1}4$ and $\cos36^\circ=\frac{\sqrt5+1}4$, with $\cos72^\circ=\sin18^\circ$. When a question mentions angles like $12^\circ$, $84^\circ$, $132^\circ$ or $\frac\pi{10}$, pair them with sum-to-product until only $18^\circ$-family angles remain.

Question 45

Choose the correct answer. $\cos2\theta\cos2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi)$ is equal to

Solution. Use the hint $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$ with $A=\theta-\phi$, $B=\theta+\phi$: $A+B=2\theta$ and $A-B=-2\phi$, so $$\sin^2(\theta-\phi)-\sin^2(\theta+\phi)=\sin2\theta\sin(-2\phi)=-\sin2\theta\sin2\phi.$$ The expression becomes $\cos2\theta\cos2\phi-\sin2\theta\sin2\phi=\cos(2\theta+2\phi)=\cos2(\theta+\phi)$.

Question 46

Choose the correct answer. The value of $\cos12^\circ+\cos84^\circ+\cos156^\circ+\cos132^\circ$ is

Solution. Pair $12^\circ$ with $132^\circ$ and $84^\circ$ with $156^\circ$ so that each half-sum is a nice angle: $$\cos12^\circ+\cos132^\circ=2\cos72^\circ\cos60^\circ=\cos72^\circ,\qquad\cos84^\circ+\cos156^\circ=2\cos120^\circ\cos36^\circ=-\cos36^\circ.$$ The total is $\cos72^\circ-\cos36^\circ=\frac{\sqrt5-1}4-\frac{\sqrt5+1}4=-\frac12$.

Question 47

Choose the correct answer. If $\tan A=\dfrac12$, $\tan B=\dfrac13$, then $\tan(2A+B)$ is equal to

Solution. First $\tan2A=\dfrac{2\cdot\frac12}{1-\frac14}=\dfrac43$. Then $$\tan(2A+B)=\frac{\frac43+\frac13}{1-\frac43\cdot\frac13}=\frac{\frac53}{\frac59}=3.$$

Question 48

Choose the correct answer. The value of $\sin\dfrac\pi{10}\sin\dfrac{13\pi}{10}$ is

Solution. $\frac\pi{10}=18^\circ$ and $\frac{13\pi}{10}=234^\circ=180^\circ+54^\circ$. So $\sin\frac{13\pi}{10}=-\sin54^\circ=-\cos36^\circ$. With the hint’s values, $$\sin18^\circ\cdot(-\cos36^\circ)=-\frac{\sqrt5-1}4\cdot\frac{\sqrt5+1}4=-\frac{5-1}{16}=-\frac14.$$

Question 49

Choose the correct answer. The value of $\sin50^\circ-\sin70^\circ+\sin10^\circ$ is equal to

Solution. Combine the two sines that sum to a nice half-angle: $\sin50^\circ+\sin10^\circ=2\sin30^\circ\cos20^\circ=\cos20^\circ=\sin70^\circ$. Hence the expression is $\sin70^\circ-\sin70^\circ=0$.

Question 50

Choose the correct answer. If $\sin\theta+\cos\theta=1$, then the value of $\sin2\theta$ is equal to

Solution. Square both sides: $\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1$, so $1+\sin2\theta=1$ and $\sin2\theta=0$.

Question 51

Choose the correct answer. If $\alpha+\beta=\dfrac\pi4$, then the value of $(1+\tan\alpha)(1+\tan\beta)$ is

Solution. From $\tan(\alpha+\beta)=1$: $\tan\alpha+\tan\beta=1-\tan\alpha\tan\beta$. Expand the product and substitute: $$(1+\tan\alpha)(1+\tan\beta)=1+(\tan\alpha+\tan\beta)+\tan\alpha\tan\beta=1+(1-\tan\alpha\tan\beta)+\tan\alpha\tan\beta=2.$$

Question 52

Choose the correct answer. If $\sin\theta=\dfrac{-4}5$ and $\theta$ lies in the third quadrant, then the value of $\cos\dfrac\theta2$ is

Solution. In the third quadrant $\cos\theta=-\sqrt{1-\frac{16}{25}}=-\frac35$. The half-angle formula gives $\cos^2\frac\theta2=\dfrac{1+\cos\theta}2=\dfrac{1-\frac35}2=\dfrac15$. For the sign: $\pi<\theta<\frac{3\pi}2$ implies $\frac\pi2<\frac\theta2<\frac{3\pi}4$, which is the second quadrant, where cosine is negative. So $\cos\frac\theta2=-\dfrac1{\sqrt5}$.

Question 53

Choose the correct answer. Number of solutions of the equation $\tan x+\sec x=2\cos x$ lying in the interval $[0,2\pi]$ is

Syllabus note: solving trigonometric equations is no longer in the current NCERT Class 11 syllabus.

Solution. Write the left side over $\cos x$ (which must be non-zero for $\tan x$ and $\sec x$ to exist): $\dfrac{\sin x+1}{\cos x}=2\cos x$, i.e. $1+\sin x=2\cos^2x=2(1-\sin x)(1+\sin x)$. Move everything to one side: $$(1+\sin x)\left[1-2(1-\sin x)\right]=0\iff(1+\sin x)(2\sin x-1)=0.$$ So $\sin x=-1$ or $\sin x=\frac12$.

• $\sin x=-1$ gives $x=\frac{3\pi}2$, but there $\cos x=0$, so $\tan x$ and $\sec x$ are undefined. Reject it.
• $\sin x=\frac12$ gives $x=\frac\pi6$ and $\frac{5\pi}6$ in $[0,2\pi]$, both with $\cos x\ne0$.

Exactly two solutions. A numerical sign-change search of $\tan x+\sec x-2\cos x$ on $[0,2\pi]$ finds the same two (it also flags $\frac{3\pi}2$, but that is a jump across the asymptote, not a root).

Question 54

Choose the correct answer. The value of $\sin\dfrac\pi{18}+\sin\dfrac\pi9+\sin\dfrac{2\pi}9+\sin\dfrac{5\pi}{18}$ is given by

Solution. In degrees the four terms are $\sin10^\circ+\sin20^\circ+\sin40^\circ+\sin50^\circ$. Pair $10^\circ$ with $50^\circ$ and $20^\circ$ with $40^\circ$, since each pair has half-sum $30^\circ$: $$\sin10^\circ+\sin50^\circ=2\sin30^\circ\cos20^\circ=\cos20^\circ=\sin70^\circ,\qquad\sin20^\circ+\sin40^\circ=2\sin30^\circ\cos10^\circ=\cos10^\circ=\sin80^\circ.$$ The total is $\sin70^\circ+\sin80^\circ=\sin\frac{7\pi}{18}+\sin\frac{4\pi}9$. Numerically the sum is $1.9245$, which rules out (B) (value $1$), (C) ($1.089$) and (D) ($1.282$).

Question 55

Choose the correct answer. If $A$ lies in the second quadrant and $3\tan A+4=0$, then the value of $2\cot A-5\cos A+\sin A$ is equal to

Solution. $\tan A=-\frac43$, so $\cot A=-\frac34$. In the second quadrant $\sin A>0$ and $\cos A<0$; the $3$-$4$-$5$ triangle gives $\sin A=\frac45$ and $\cos A=-\frac35$. Then $$2\cot A-5\cos A+\sin A=-\frac32+3+\frac45=\frac{-15+30+8}{10}=\frac{23}{10}.$$

Question 56

Choose the correct answer. The value of $\cos^248^\circ-\sin^212^\circ$ is

Solution. Use the hint $\cos^2A-\sin^2B=\cos(A+B)\cos(A-B)$ with $A=48^\circ$, $B=12^\circ$: $$\cos60^\circ\cos36^\circ=\frac12\cdot\frac{\sqrt5+1}4=\frac{\sqrt5+1}8.$$

Question 57

Choose the correct answer. If $\tan\alpha=\dfrac17$, $\tan\beta=\dfrac13$, then $\cos2\alpha$ is equal to

Solution. Compute each side with the $\tan$ form of the double-angle formulae. First $\cos2\alpha=\dfrac{1-\frac1{49}}{1+\frac1{49}}=\dfrac{48}{50}=\dfrac{24}{25}$. Next, $\tan2\beta=\dfrac{\frac23}{1-\frac19}=\dfrac34$, and $\sin4\beta=\dfrac{2\tan2\beta}{1+\tan^22\beta}=\dfrac{\frac32}{\frac{25}{16}}=\dfrac{24}{25}$. They agree. The other options are different numbers: $\sin2\beta=\frac35$, $\cos2\beta=\frac45$, $\sin3\beta=\frac{26}{10\sqrt{10}}\approx0.822$, none of which equals $\frac{24}{25}=0.96$.

Question 58

Choose the correct answer. If $\tan\theta=\dfrac ab$, then $b\cos2\theta+a\sin2\theta$ is equal to

Solution. Write both double angles in terms of $t=\tan\theta=\frac ab$: $\cos2\theta=\frac{1-t^2}{1+t^2}=\frac{b^2-a^2}{a^2+b^2}$ and $\sin2\theta=\frac{2t}{1+t^2}=\frac{2ab}{a^2+b^2}$. Then $$b\cos2\theta+a\sin2\theta=\frac{b(b^2-a^2)+2a^2b}{a^2+b^2}=\frac{b(a^2+b^2)}{a^2+b^2}=b.$$

Question 59

Choose the correct answer. If for real values of $x$, $\cos\theta=x+\dfrac1x$, then

Solution. For real $x\neq0$, $\left|x+\frac1x\right|\ge2$: for $x>0$ this is AM-GM, and for $x<0$ apply it to $-x$. But $|\cos\theta|\le1$, so the equation $\cos\theta=x+\frac1x$ has no solution for any real $\theta$.

Common mistakes

• Question 46: pairing $12^\circ$ with $84^\circ$. The useful pairs are the ones whose half-sums are $72^\circ$ and $120^\circ$; wrong pairing leaves angles with no known exact value.
• Question 47: using $\tan(A+B+A)$ directly. Compute $\tan2A$ first, then add $B$.
• Question 48: forgetting that $\sin\frac{13\pi}{10}$ is negative (the angle is in the third quadrant). Dropping the sign gives $+\frac14$, which is not among the options and should prompt a recheck.
• Question 52: taking $\cos\frac\theta2$ positive. The half-angle $\frac\theta2$ lies in the second quadrant, so the square root must be negative.
• Question 53: counting $x=\frac{3\pi}2$ as a third root. It satisfies the cleared equation $(1+\sin x)(2\sin x-1)=0$ but is not in the domain of $\tan x$ and $\sec x$.
• Question 59: concluding $\theta$ is obtuse or acute from the sign of $x+\frac1x$. The magnitude test $|x+\frac1x|\ge2$ settles it before any sign question arises.

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 30 to 44 and Fill in the Blanks, True or False and Matching, questions 60 to 76.
• NCERT textbook: Exercise 3.3 (sum-to-product and half-angle proofs), Exercise 3.2 and the Miscellaneous Exercise.