NCERT Exemplar Solutions Class 11 Maths Trigonometric Functions — Short Answer Type Q1–19

NCERT Exemplar Problems, Class 11 Mathematics — Trigonometric Functions, Short Answer Type questions 1 to 19. All 19 questions solved.

Nineteen short-answer questions. Questions 1 to 14 are identity proofs and evaluations that run on four tools: the compound-angle formulae $\sin(A\pm B)$, $\cos(A\pm B)$, $\tan(A\pm B)$; the double-angle formulae; the sum-to-product formulae; and componendo and dividendo. Questions 15 to 19 solve trigonometric equations.

Two things in the printed material need correcting. Question 6 has a misprint in its right-hand side (the identity as printed is false), and the printed key for Question 19 includes $x=\frac\pi2$, which is not a solution. Both are explained where they occur.

Key insight. In an identity proof, work on one side only and look for the move that turns a sum into a product or a ratio into a single function: a half-angle $\tan$, a compound angle, a common factor. In a conditional proof, rewrite the given relation as a ratio and apply componendo and dividendo, so that the sum and difference collapse into product formulae.

Question 1

Prove that $\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1+\sin A}{\cos A}$.

Solution. The numerator hides a factor. Use $\sec^2A-\tan^2A=1$ to replace the $-1$ by $-(\sec A-\tan A)(\sec A+\tan A)$. Then $$\tan A+\sec A-1=(\tan A+\sec A)-(\sec A+\tan A)(\sec A-\tan A)=(\tan A+\sec A)(1-\sec A+\tan A).$$ The second bracket is exactly the denominator $\tan A-\sec A+1$, so it cancels (it is non-zero wherever the left-hand side is defined): $$\text{LHS}=\tan A+\sec A=\frac{\sin A}{\cos A}+\frac{1}{\cos A}=\frac{1+\sin A}{\cos A}=\text{RHS}.$$

Question 2

If $\dfrac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}=y$, then prove that $\dfrac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}$ is also equal to $y$.

Solution. Follow the hint: multiply and divide by $1+\cos\alpha+\sin\alpha$, so that the numerator becomes a difference of squares. $$\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}\cdot\frac{1+\cos\alpha+\sin\alpha}{1+\cos\alpha+\sin\alpha}=\frac{(1+\sin\alpha)^2-\cos^2\alpha}{(1+\sin\alpha)(1+\cos\alpha+\sin\alpha)}.$$ Expand the numerator and replace $\cos^2\alpha$ by $1-\sin^2\alpha$: $$(1+\sin\alpha)^2-\cos^2\alpha=1+2\sin\alpha+\sin^2\alpha-1+\sin^2\alpha=2\sin\alpha(1+\sin\alpha).$$ Cancel $1+\sin\alpha$ to get $\dfrac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}=y$.

Question 3

If $m\sin\theta=n\sin(\theta+2\alpha)$, then prove that $\tan(\theta+\alpha)\cot\alpha=\dfrac{m+n}{m-n}$.

Solution. The given relation says $\dfrac{\sin(\theta+2\alpha)}{\sin\theta}=\dfrac mn$. Componendo and dividendo turns a ratio $\frac pq=\frac mn$ into $\frac{p+q}{p-q}=\frac{m+n}{m-n}$, so $$\frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\alpha)-\sin\theta}=\frac{m+n}{m-n}.$$ Use the sum-to-product formulae $\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$ and $\sin C-\sin D=2\cos\frac{C+D}2\sin\frac{C-D}2$ with $C=\theta+2\alpha$, $D=\theta$. The half-sum is $\theta+\alpha$ and the half-difference is $\alpha$: $$\frac{2\sin(\theta+\alpha)\cos\alpha}{2\cos(\theta+\alpha)\sin\alpha}=\tan(\theta+\alpha)\cot\alpha=\frac{m+n}{m-n}.$$

Question 4

If $\cos(\alpha+\beta)=\dfrac45$ and $\sin(\alpha-\beta)=\dfrac5{13}$, where $\alpha$ lies between $0$ and $\dfrac\pi4$, find the value of $\tan2\alpha$.

Solution. Write $2\alpha=(\alpha+\beta)+(\alpha-\beta)$, so that $\tan2\alpha$ needs only $\tan(\alpha+\beta)$ and $\tan(\alpha-\beta)$. From $\cos(\alpha+\beta)=\frac45$ we get $\sin(\alpha+\beta)=\frac35$ and $\tan(\alpha+\beta)=\frac34$. From $\sin(\alpha-\beta)=\frac5{13}$ we get $\cos(\alpha-\beta)=\frac{12}{13}$ and $\tan(\alpha-\beta)=\frac5{12}$.

The signs of the missing sine and cosine are not stated, but the condition $0<\alpha<\frac\pi4$ fixes them: it puts $2\alpha$ in $\left(0,\frac\pi2\right)$. Of the four possible sign combinations, only taking $\sin(\alpha+\beta)$ and $\cos(\alpha-\beta)$ both positive gives $2\alpha\approx0.6435+0.3948\approx1.04$ radians, which lies in $\left(0,\frac\pi2\right)$; the other three put $2\alpha$ outside that interval.

$$\tan2\alpha=\frac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta)\tan(\alpha-\beta)}=\frac{\frac34+\frac5{12}}{1-\frac34\cdot\frac5{12}}=\frac{\frac{14}{12}}{\frac{33}{48}}=\frac{56}{33}.$$

Question 5

If $\tan x=\dfrac ba$, then find the value of $\sqrt{\dfrac{a+b}{a-b}}+\sqrt{\dfrac{a-b}{a+b}}$.

Solution. Put the two roots over a common denominator (taking $a>0$ and $a>|b|$ so that both radicals are real and positive): $$\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{(a+b)+(a-b)}{\sqrt{(a-b)(a+b)}}=\frac{2a}{\sqrt{a^2-b^2}}.$$ In terms of $x$, divide top and bottom by $a$ to get $\dfrac{2}{\sqrt{1-\tan^2x}}$. Since $1-\tan^2x=\dfrac{\cos2x}{\cos^2x}$, this is $\dfrac{2\cos x}{\sqrt{\cos2x}}$, which is the form the printed key gives.

Question 6

Prove that $\cos\theta\cos\dfrac\theta2-\cos3\theta\cos\dfrac{9\theta}2=\sin7\theta\sin8\theta$ (as printed).

The identity as printed is false. At $\theta=0.37$ the left side is $0.958$ and $\sin7\theta\sin8\theta$ is $0.095$. The right-hand side has lost its fractions: the correct identity is $$\cos\theta\cos\frac\theta2-\cos3\theta\cos\frac{9\theta}2=\sin\frac{7\theta}2\sin4\theta,$$ which is what the hint’s working actually produces. We prove that version.

Solution. Follow the hint and turn each product into a sum with $2\cos P\cos Q=\cos(P+Q)+\cos(P-Q)$: $$\text{LHS}=\tfrac12\left[2\cos\theta\cos\tfrac\theta2-2\cos3\theta\cos\tfrac{9\theta}2\right]=\tfrac12\left[\cos\tfrac{3\theta}2+\cos\tfrac\theta2-\cos\tfrac{15\theta}2-\cos\tfrac{3\theta}2\right].$$ The $\cos\frac{3\theta}2$ terms cancel, leaving $\tfrac12\left[\cos\frac\theta2-\cos\frac{15\theta}2\right]$. Now $\cos C-\cos D=2\sin\frac{C+D}2\sin\frac{D-C}2$, with half-sum $4\theta$ and half-difference $\frac{7\theta}2$: $$\text{LHS}=\sin4\theta\,\sin\frac{7\theta}2.$$

Question 7

If $a\cos\theta+b\sin\theta=m$ and $a\sin\theta-b\cos\theta=n$, then show that $a^2+b^2=m^2+n^2$.

Solution. Square both given equations and add, so that the cross terms cancel: $$m^2+n^2=(a^2\cos^2\theta+2ab\sin\theta\cos\theta+b^2\sin^2\theta)+(a^2\sin^2\theta-2ab\sin\theta\cos\theta+b^2\cos^2\theta).$$ The $2ab\sin\theta\cos\theta$ terms cancel and $\sin^2\theta+\cos^2\theta=1$ takes care of the rest: $m^2+n^2=a^2+b^2$.

Question 8

Find the value of $\tan22^\circ30’$.

Solution. $22^\circ30’=\frac12\cdot45^\circ$, so use the hint with $\theta=45^\circ$: $\tan\dfrac\theta2=\dfrac{\sin\theta}{1+\cos\theta}$. This holds because $\dfrac{2\sin\frac\theta2\cos\frac\theta2}{2\cos^2\frac\theta2}$ simplifies to the left side. $$\tan22^\circ30’=\frac{\sin45^\circ}{1+\cos45^\circ}=\frac{\frac1{\sqrt2}}{1+\frac1{\sqrt2}}=\frac1{\sqrt2+1}=\sqrt2-1.$$

Question 9

Prove that $\sin4A=4\sin A\cos^3A-4\cos A\sin^3A$.

Solution. Apply the double-angle formula twice, since $4A=2(2A)$: $$\sin4A=2\sin2A\cos2A=2(2\sin A\cos A)(\cos^2A-\sin^2A)=4\sin A\cos^3A-4\sin^3A\cos A.$$

Question 10

If $\tan\theta+\sin\theta=m$ and $\tan\theta-\sin\theta=n$, then prove that $m^2-n^2=4\sin\theta\tan\theta$.

Solution. The target is a difference of squares, so factor it: $m^2-n^2=(m+n)(m-n)$. Adding the two equations gives $m+n=2\tan\theta$; subtracting gives $m-n=2\sin\theta$. Hence $m^2-n^2=2\tan\theta\cdot2\sin\theta=4\sin\theta\tan\theta$.

Question 11

If $\tan(A+B)=p$, $\tan(A-B)=q$, then show that $\tan2A=\dfrac{p+q}{1-pq}$.

Solution. Write $2A=(A+B)+(A-B)$ and use the formula for $\tan(X+Y)$ with $X=A+B$, $Y=A-B$: $$\tan2A=\frac{\tan(A+B)+\tan(A-B)}{1-\tan(A+B)\tan(A-B)}=\frac{p+q}{1-pq}.$$

Question 12

If $\cos\alpha+\cos\beta=0=\sin\alpha+\sin\beta$, then prove that $\cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta)$.

Solution. The hint subtracts the squares of the two given sums: $(\cos\alpha+\cos\beta)^2-(\sin\alpha+\sin\beta)^2=0-0=0$. Expand and regroup: $$(\cos^2\alpha-\sin^2\alpha)+(\cos^2\beta-\sin^2\beta)+2(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=0,$$ that is $\cos2\alpha+\cos2\beta+2\cos(\alpha+\beta)=0$. Moving the last term across gives the result.

Question 13

If $\dfrac{\sin(x+y)}{\sin(x-y)}=\dfrac{a+b}{a-b}$, then show that $\dfrac{\tan x}{\tan y}=\dfrac ab$.

Solution. Apply componendo and dividendo to the given ratio: $$\frac{\sin(x+y)+\sin(x-y)}{\sin(x+y)-\sin(x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{2a}{2b}=\frac ab.$$ Expanding $\sin(x\pm y)$ makes the left side easy: the numerator is $2\sin x\cos y$ and the denominator is $2\cos x\sin y$. So the left side is $\dfrac{\sin x\cos y}{\cos x\sin y}=\dfrac{\tan x}{\tan y}$.

Question 14

If $\tan\theta=\dfrac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$, then show that $\sin\alpha+\cos\alpha=\sqrt2\cos\theta$.

Solution. Divide the numerator and denominator of the right side by $\cos\alpha$: $$\tan\theta=\frac{\tan\alpha-1}{\tan\alpha+1}=\frac{\tan\alpha-\tan\frac\pi4}{1+\tan\alpha\tan\frac\pi4}=\tan\left(\alpha-\frac\pi4\right),$$ so $\theta=\alpha-\frac\pi4$. This is the choice the hint makes; adding a multiple of $\pi$ to $\theta$ would flip the sign of $\cos\theta$ for odd multiples, so the identity is understood for this principal choice. Then $\alpha=\theta+\frac\pi4$ and $$\sin\alpha+\cos\alpha=\sqrt2\left(\sin\alpha\cos\tfrac\pi4+\cos\alpha\sin\tfrac\pi4\right)=\sqrt2\sin\left(\alpha+\tfrac\pi4\right)=\sqrt2\sin\left(\theta+\tfrac\pi2\right)=\sqrt2\cos\theta.$$

Question 15

If $\sin\theta+\cos\theta=1$, then find the general value of $\theta$.

Syllabus note: general solutions of trigonometric equations are no longer in the current NCERT Class 11 syllabus.

Solution. Write $\sin\theta+\cos\theta=\sqrt2\sin\left(\theta+\frac\pi4\right)$, so the equation is $\sin\left(\theta+\frac\pi4\right)=\frac1{\sqrt2}=\sin\frac\pi4$. The general solution of $\sin u=\sin v$ is $u=n\pi+(-1)^nv$, so $$\theta+\frac\pi4=n\pi+(-1)^n\frac\pi4\quad\Rightarrow\quad\theta=n\pi+(-1)^n\frac\pi4-\frac\pi4,\ n\in\mathbb Z.$$ Splitting by parity gives $\theta=2k\pi$ (for $n=2k$) or $\theta=2k\pi+\frac\pi2$ (for $n=2k+1$). A direct check agrees: squaring gives $\sin2\theta=0$, so $\theta$ is a multiple of $\frac\pi2$, and of those only $0$ and $\frac\pi2$ (plus full turns) satisfy the original equation; $\pi$ and $\frac{3\pi}2$ give $-1$.

Question 16

Find the most general value of $\theta$ satisfying the equation $\tan\theta=-1$ and $\cos\theta=\dfrac1{\sqrt2}$.

Syllabus note: general solutions of trigonometric equations are no longer in the current NCERT Class 11 syllabus.

Solution. $\tan\theta=-1$ puts $\theta$ in the second or fourth quadrant; $\cos\theta>0$ puts it in the first or fourth. Both hold only in the fourth quadrant, where the reference angle is $\frac\pi4$, so $\theta=2\pi-\frac\pi4=\frac{7\pi}4$. Adding full turns gives every such angle.

Question 17

If $\cot\theta+\tan\theta=2\,\mathrm{cosec}\,\theta$, then find the general value of $\theta$.

Syllabus note: general solutions of trigonometric equations are no longer in the current NCERT Class 11 syllabus.

Solution. Combine the left side: $\cot\theta+\tan\theta=\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\dfrac1{\sin\theta\cos\theta}$. The equation becomes $\dfrac1{\sin\theta\cos\theta}=\dfrac2{\sin\theta}$. Since $\sin\theta\neq0$ (otherwise $\mathrm{cosec}\,\theta$ and $\cot\theta$ are undefined), cancel it: $\cos\theta=\frac12$. The general solution of $\cos\theta=\cos\frac\pi3$ is $\theta=2n\pi\pm\frac\pi3$.

Question 18

If $2\sin^2\theta=3\cos\theta$, where $0\le\theta\le2\pi$, then find the value of $\theta$.

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

Solution. Convert to a single function with $\sin^2\theta=1-\cos^2\theta$: $2-2\cos^2\theta=3\cos\theta$, i.e. $2\cos^2\theta+3\cos\theta-2=0$. This factorises as $(2\cos\theta-1)(\cos\theta+2)=0$. The factor $\cos\theta+2$ is never zero, since $\cos\theta\ge-1$. So $\cos\theta=\frac12$, whose solutions in $[0,2\pi]$ are $\frac\pi3$ and $2\pi-\frac\pi3=\frac{5\pi}3$.

Question 19

If $\sec x\cos5x+1=0$, where $0<x\le\dfrac\pi2$, then find the value of $x$.

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

Solution. Multiply by $\cos x$ (allowed because $\sec x$ must be defined, so $\cos x\ne0$): $\cos5x+\cos x=0$. Apply sum-to-product: $2\cos3x\cos2x=0$.

• $\cos3x=0$ gives $3x=\frac\pi2,\frac{3\pi}2,\dots$, so in $\left(0,\frac\pi2\right]$ the values are $x=\frac\pi6$ and $x=\frac\pi2$.
• $\cos2x=0$ gives $2x=\frac\pi2$, so $x=\frac\pi4$ ($2x=\frac{3\pi}2$ would give $x=\frac{3\pi}4$, outside the interval).

Now test each candidate in the original equation. $x=\frac\pi2$ fails at once: $\sec\frac\pi2$ does not exist, so $\frac\pi2$ is an extraneous root brought in by multiplying through by $\cos x$. The others check: $x=\frac\pi6$ gives $\frac2{\sqrt3}\cdot\left(-\frac{\sqrt3}2\right)+1=0$, and $x=\frac\pi4$ gives $\sqrt2\cdot\left(-\frac1{\sqrt2}\right)+1=0$. A numerical search for sign changes of $\frac{\cos5x}{\cos x}+1$ on $\left(0,\frac\pi2\right)$ finds exactly these two roots.

The printed key lists $\frac\pi6,\frac\pi4,\frac\pi2$; the last is wrong for the reason above.

Common mistakes

• Questions 1 and 2: expanding both sides fully instead of looking for the factor. In Question 1 the numerator factors once $1$ is written as $\sec^2A-\tan^2A$; in Question 2 multiplying by the conjugate gives a difference of squares. Brute-force expansion works but invites sign errors.
• Question 4: assuming the signs of $\sin(\alpha+\beta)$ and $\cos(\alpha-\beta)$ without checking. The condition on $\alpha$ is what fixes them, and skipping that check is how students pick up a wrong sign in the final fraction.
• Question 6: forcing the printed right-hand side. The question is misprinted, so students who keep hunting for their own error lose time. When an identity will not come out, test one numerical value first.
• Question 14: mixing up $\sin\left(\alpha+\frac\pi4\right)$ with $\cos\left(\alpha-\frac\pi4\right)$. They are equal because $\sin\left(u+\frac\pi2\right)=\cos u$ with $u=\alpha-\frac\pi4$, but it is easy to lose the $\frac\pi2$ shift.
• Question 15: squaring the equation and not testing the roots. Squaring creates the extra roots $\pi$ and $\frac{3\pi}2$, which actually give $\sin\theta+\cos\theta=-1$.
• Questions 17 and 19: cancelling $\sin\theta$ or multiplying by $\cos x$ without checking where the original expression is defined. In Question 19 this costs the answer: $x=\frac\pi2$ comes out of $\cos3x=0$ but makes $\sec x$ undefined.

Practise next

• The same unit’s other exemplar pages: Long Answer Type, questions 20 to 29, Objective Type, questions 30 to 44, Objective Type, questions 45 to 59 and Fill in the Blanks, True or False and Matching, questions 60 to 76.
• NCERT textbook: Exercise 3.3 (compound angles, sum-to-product and multiple-angle proofs), Exercise 3.2 and the Miscellaneous Exercise.