Introduction to Trigonometry Class 10 Practice Questions

Ques­tion 2#

In $\triangle PQR$, right-angled at $Q$, $PQ = 9$ cm and $QR = 12$ cm. Find the value of $\sin R \cos P + \cos R \sin P$.

Solu­tion. $PR = \sqrt{81 + 144} = \sqrt{225} = 15$ cm.

For $\angle P$: oppo­site $= QR = 12$, adja­cent $= PQ = 9$, so $\displaystyle \sin P = \frac{12}{15} = \frac{4}{5}$ and $\displaystyle \cos P = \frac{9}{15} = \frac{3}{5}$.

For $\angle R$: oppo­site $= PQ = 9$, adja­cent $= QR = 12$, so $\displaystyle \sin R = \frac{3}{5}$ and $\displaystyle \cos R = \frac{4}{5}$.

$$\displaystyle \sin R \cos P + \cos R \sin P = \frac{3}{5} \cdot \frac{3}{5} + \frac{4}{5} \cdot \frac{4}{5} = \frac{9 + 16}{25} = 1.$$

Answer: $1$.

Why this works: $P$ and $R$ are the two acute angles of the same right tri­an­gle, so the side oppo­site one is adja­cent to the other; hence $\sin R = \cos P$ and $\cos R = \sin P$.

Ques­tion 3#

In $\triangle XYZ$, right-angled at $Y$, $XY = 7$ cm and $XZ - YZ = 1$ cm. Find $\sin X$ and $\tan Z$.

Solu­tion. By Pythago­ras' the­o­rem, $XZ^2 - YZ^2 = XY^2 = 49$. Fac­toris­ing,

$$(XZ - YZ)(XZ + YZ) = 49 \;\Rightarrow\; 1 \times (XZ + YZ) = 49 \;\Rightarrow\; XZ + YZ = 49.$$

Adding the two equa­tions: $2\,XZ = 50$, so $XZ = 25$ cm and $YZ = 24$ cm.

$$\displaystyle \sin X = \frac{YZ}{XZ} = \frac{24}{25}, \qquad \tan Z = \frac{XY}{YZ} = \frac{7}{24}.$$

Answer: $\displaystyle \sin X = \frac{24}{25}$, $\displaystyle \tan Z = \frac{7}{24}$.

From one ratio to all the oth­ers#

Ques­tion 4#

If $\displaystyle \cos A = \frac{20}{29}$, find the other five trigono­met­ric ratios of $A$.

Solu­tion. Take adja­cent $= 20k$ and hypotenuse $= 29k$. Then

$$\text{opposite} = \sqrt{(29k)^2 - (20k)^2} = \sqrt{841k^2 - 400k^2} = \sqrt{441k^2} = 21k.$$

$$\displaystyle \sin A = \frac{21}{29},\quad \tan A = \frac{21}{20},\quad \operatorname{cosec} A = \frac{29}{21},\quad \sec A = \frac{29}{20},\quad \cot A = \frac{20}{21}.$$

Answer: $\displaystyle \sin A = \frac{21}{29}$, $\displaystyle \tan A = \frac{21}{20}$, $\displaystyle \operatorname{cosec} A = \frac{29}{21}$, $\displaystyle \sec A = \frac{29}{20}$, $\displaystyle \cot A = \frac{20}{21}$.

Ques­tion 5#

If $5 \tan \theta = 12$, eval­u­ate $\displaystyle \frac{2\sin \theta - \cos \theta}{\sin \theta + 3\cos \theta}$.

Solu­tion. $\displaystyle \tan \theta = \frac{12}{5}$, so the sides are $12k$ and $5k$, and the hypotenuse is $\sqrt{144k^2 + 25k^2} = 13k$. Hence $\displaystyle \sin \theta = \frac{12}{13}$ and $\displaystyle \cos \theta = \frac{5}{13}$.

$$\displaystyle \frac{2 \cdot \frac{12}{13} - \frac{5}{13}}{\frac{12}{13} + 3 \cdot \frac{5}{13}} = \frac{\frac{24 - 5}{13}}{\frac{12 + 15}{13}} = \frac{19}{27}.$$

Answer: $\displaystyle \frac{19}{27}$.

Why this works: you could also divide the numer­a­tor and denom­i­na­tor by $\cos \theta$ to get $\displaystyle \frac{2\tan\theta - 1}{\tan\theta + 3} = \frac{24/5 - 1}{12/5 + 3} = \frac{19}{27}$, with no tri­an­gle needed.

Ques­tion 6#

If $A$ is acute and $\displaystyle \sin A = \frac{1}{\sqrt{10}}$, find $\cos A$ and $\tan A$, and eval­u­ate $3\cos^2 A - \sin^2 A$.

Solu­tion. Using $\sin^2 A + \cos^2 A = 1$:

$$\displaystyle \cos^2 A = 1 - \frac{1}{10} = \frac{9}{10}, \qquad \cos A = \frac{3}{\sqrt{10}}, \qquad \tan A = \frac{\sin A}{\cos A} = \frac{1}{3}.$$

$$\displaystyle 3\cos^2 A - \sin^2 A = 3 \cdot \frac{9}{10} - \frac{1}{10} = \frac{26}{10} = \frac{13}{5}.$$

Answer: $\displaystyle \cos A = \frac{3}{\sqrt{10}}$, $\displaystyle \tan A = \frac{1}{3}$, and the value is $\displaystyle \frac{13}{5}$.

Eval­u­at­ing at stan­dard angles#

Ques­tion 7#

Eval­u­ate $3\tan^2 30^\circ + 2\sin 30^\circ \cos 60^\circ - \cot^2 45^\circ$.

Solu­tion.

$$\displaystyle 3 \cdot \frac{1}{3} + 2 \cdot \frac{1}{2} \cdot \frac{1}{2} - 1^2 = 1 + \frac{1}{2} - 1 = \frac{1}{2}.$$

Answer: $\displaystyle \frac{1}{2}$.

Ques­tion 8#

Eval­u­ate $\displaystyle \frac{4\cos^2 45^\circ + 3\cot^2 60^\circ - \sec^2 60^\circ}{\tan^2 60^\circ - \operatorname{cosec}^2 30^\circ}$.

Solu­tion. Numer­a­tor:

$$\displaystyle 4 \cdot \frac{1}{2} + 3 \cdot \frac{1}{3} - 2^2 = 2 + 1 - 4 = -1.$$

Denom­i­na­tor: $(\sqrt{3})^2 - 2^2 = 3 - 4 = -1$.

So the value is $\displaystyle \frac{-1}{-1} = 1$.

Answer: $1$.

Ques­tion 9#

Eval­u­ate $\displaystyle \frac{\tan 45^\circ + \sin 60^\circ}{\operatorname{cosec} 30^\circ - \cos 30^\circ}$ and express the answer with a ratio­nal denom­i­na­tor.

Solu­tion.

$$\displaystyle \begin{aligned} \frac{1 + \frac{\sqrt{3}}{2}}{2 - \frac{\sqrt{3}}{2}} &= \frac{2 + \sqrt{3}}{4 - \sqrt{3}} = \frac{(2 + \sqrt{3})(4 + \sqrt{3})}{(4 - \sqrt{3})(4 + \sqrt{3})} \\ &= \frac{8 + 2\sqrt{3} + 4\sqrt{3} + 3}{16 - 3} = \frac{11 + 6\sqrt{3}}{13}. \end{aligned}$$

Answer: $\displaystyle \frac{11 + 6\sqrt{3}}{13}$.

Find­ing unknown angles#

Ques­tion 10#

Find $\theta$ if $2\sin (2\theta - 10^\circ) = \sqrt{3}$, where $2\theta - 10^\circ$ is an acute angle.

Solu­tion. $\displaystyle \sin (2\theta - 10^\circ) = \frac{\sqrt{3}}{2} = \sin 60^\circ$. So

$$2\theta - 10^\circ = 60^\circ \;\Rightarrow\; 2\theta = 70^\circ \;\Rightarrow\; \theta = 35^\circ.$$

Answer: $\theta = 35^\circ$.

Ques­tion 11#

If $\sin (A + B) = 1$ and $\displaystyle \tan (A - B) = \frac{1}{\sqrt{3}}$, where $0^\circ \lt A + B \le 90^\circ$ and $A \gt B$, find $A$ and $B$.

Solu­tion. $\sin 90^\circ = 1$, so $A + B = 90^\circ$. $\displaystyle \tan 30^\circ = \frac{1}{\sqrt{3}}$, so $A - B = 30^\circ$.

Adding: $2A = 120^\circ$, so $A = 60^\circ$. Then $B = 90^\circ - 60^\circ = 30^\circ$.

Check: $\sin 90^\circ = 1$ and $\displaystyle \tan 30^\circ = \frac{1}{\sqrt{3}}$.

Answer: $A = 60^\circ$, $B = 30^\circ$.

Sim­pli­fi­ca­tion and iden­tity proofs#

Ques­tion 12#

Sim­plify $(1 + \cot^2 A)(1 - \cos A)(1 + \cos A)$.

Solu­tion.

$$\displaystyle (1 + \cot^2 A)(1 - \cos^2 A) = \operatorname{cosec}^2 A \cdot \sin^2 A = \frac{1}{\sin^2 A} \cdot \sin^2 A = 1.$$

Answer: $1$.

Ques­tion 13#

Prove that $\displaystyle \frac{\cos \theta}{1 - \sin \theta} - \frac{\cos \theta}{1 + \sin \theta} = 2\tan \theta$.

Solu­tion. Take the com­mon denom­i­na­tor $(1 - \sin \theta)(1 + \sin \theta) = 1 - \sin^2 \theta$:

$$\displaystyle \begin{aligned} \text{LHS} &= \frac{\cos \theta\,(1 + \sin \theta) - \cos \theta\,(1 - \sin \theta)}{1 - \sin^2 \theta} = \frac{2\sin \theta \cos \theta}{\cos^2 \theta} \\ &= \frac{2\sin \theta}{\cos \theta} = 2\tan \theta = \text{RHS}. \end{aligned}$$

The steps are valid for $0^\circ \le \theta \lt 90^\circ$: then $1 - \sin \theta \ne 0$ and $\cos \theta \ne 0$, so the can­cel­la­tion of $\cos \theta$ is allowed.

Answer: LHS = RHS, so the iden­tity is proved.

Ques­tion 14#

Prove that $(\sin A + \cos A)^2 + (\sin A - \cos A)^2 = 2$.

Solu­tion. Expand both squares:

$$\begin{aligned} \text{LHS} &= (\sin^2 A + 2\sin A\cos A + \cos^2 A) + (\sin^2 A - 2\sin A\cos A + \cos^2 A) \\ &= 2(\sin^2 A + \cos^2 A) = 2 \times 1 = 2 = \text{RHS}. \end{aligned}$$

Answer: LHS = RHS, so the iden­tity is proved.

Ques­tion 15#

Prove that $\displaystyle \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{1 + \sin A}{\cos A}$, where $0^\circ \lt A \lt 90^\circ$.

Solu­tion. Divide the numer­a­tor and the denom­i­na­tor by $\cos A$ (which is not zero):

$$\displaystyle \text{LHS} = \frac{\tan A - 1 + \sec A}{\tan A + 1 - \sec A} = \frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}.$$

Now replace the 1 in the numer­a­tor by $\sec^2 A - \tan^2 A$:

$$\displaystyle \begin{aligned} \text{LHS} &= \frac{(\tan A + \sec A) - (\sec^2 A - \tan^2 A)}{\tan A - \sec A + 1} \\ &= \frac{(\sec A + \tan A) - (\sec A - \tan A)(\sec A + \tan A)}{\tan A - \sec A + 1} \\ &= \frac{(\sec A + \tan A)(1 - \sec A + \tan A)}{\tan A - \sec A + 1}. \end{aligned}$$

The bracket $1 - \sec A + \tan A$ is the same as the denom­i­na­tor, so it can­cels:

$$\displaystyle \text{LHS} = \sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A} = \text{RHS}.$$

Answer: LHS = RHS, so the iden­tity is proved.

Why this works: writ­ing 1 as $\sec^2 A - \tan^2 A$ cre­ates a com­mon fac­tor $\sec A + \tan A$; this is the stan­dard trick for expres­sions with a stray 1.

Ques­tion 16#

If $\operatorname{cosec} \theta + \cot \theta = 5$, find $\operatorname{cosec} \theta - \cot \theta$ and hence $\cos \theta$.

Solu­tion. Since $\operatorname{cosec}^2 \theta - \cot^2 \theta = 1$,

$$\displaystyle (\operatorname{cosec} \theta - \cot \theta)(\operatorname{cosec} \theta + \cot \theta) = 1 \;\Rightarrow\; \operatorname{cosec} \theta - \cot \theta = \frac{1}{5}.$$

Adding and sub­tract­ing the two equa­tions:

$$\displaystyle 2\operatorname{cosec} \theta = 5 + \frac{1}{5} = \frac{26}{5} \;\Rightarrow\; \operatorname{cosec} \theta = \frac{13}{5}, \qquad 2\cot \theta = 5 - \frac{1}{5} = \frac{24}{5} \;\Rightarrow\; \cot \theta = \frac{12}{5}.$$

$$\displaystyle \cos \theta = \frac{\cot \theta}{\operatorname{cosec} \theta} = \frac{12/5}{13/5} = \frac{12}{13}.$$

Answer: $\displaystyle \operatorname{cosec} \theta - \cot \theta = \frac{1}{5}$, $\displaystyle \cos \theta = \frac{12}{13}$.

True or false with rea­sons#

Ques­tion 17#

State whether each state­ment is true or false, and jus­tify your answer.

1. The value of $\cot A$ can never be greater than 1.
2. $\displaystyle \operatorname{cosec} A = \frac{7}{4}$ for some angle $A$.
3. $\displaystyle \sec A = \frac{3}{4}$ for some angle $A$.
4. $\sin A = \cos A$ when $A = 45^\circ$.
5. $\sin A + \cos A = 1$ for every angle $A$.

Solu­tion.

1. False. $\cot 30^\circ = \sqrt{3} \approx 1.73$, which is greater than 1.
2. True. $\operatorname{cosec} A \ge 1$, and $\displaystyle \frac{7}{4} = 1.75 \gt 1$. Such an angle exists: a right tri­an­gle with hypotenuse 7 and oppo­site side 4 is pos­si­ble (the third side is $\sqrt{33}$).
3. False. $\displaystyle \sec A = \frac{\text{hypotenuse}}{\text{adjacent}}$, and the hypotenuse can­not be shorter than the adja­cent side, so $\sec A \ge 1$; but $\displaystyle \frac{3}{4} \lt 1$.
4. True. $\displaystyle \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$.
5. False. At $A = 45^\circ$, $\displaystyle \sin A + \cos A = \frac{2}{\sqrt{2}} = \sqrt{2} \ne 1$. The cor­rect iden­tity is $\sin^2 A + \cos^2 A = 1$.

Answer: (1) False, (2) True, (3) False, (4) True, (5) False.

Key terms#

Trigono­met­ric ratio

A ratio of two sides of a right tri­an­gle, which depends only on the acute angle cho­sen.

Hypotenuse

The side oppo­site the right angle, and the longest side of the tri­an­gle.

Stan­dard angles

0°, 30°, 45°, 60° and 90°, whose ratios have exact val­ues.

Iden­tity

An equa­tion that holds for every angle for which both sides are defined.

Rec­i­p­ro­cal rela­tion

A link such as sec A = 1/cos A between a ratio and its rec­i­p­ro­cal ratio.

Ratio­nal denom­i­na­tor

A denom­i­na­tor with no surd, obtained by mul­ti­ply­ing by a suit­able con­ju­gate.

Not defined

Describes a ratio that would need divi­sion by zero, such as tan 90°.

Com­mon ques­tions#

How many marks do iden­tity proofs usu­ally carry?#

In board papers, a sin­gle iden­tity proof is usu­ally a 2-mark or 3-mark ques­tion, and a longer one may be part of a larger ques­tion. Show each step on its own line.

Should I draw a tri­an­gle for every "one ratio to all" ques­tion?#

A quick sketch helps you avoid mix­ing up oppo­site and adja­cent sides. For sim­ple cases, the iden­ti­ties give the answer just as quickly.

What if my iden­tity proof gets stuck halfway?#

Go back and con­vert every­thing to sine and cosine, or sim­plify the other side sep­a­rately. If both sides reach the same expres­sion, the proof is com­plete.

Do I need to state the range of the angle in my answer?#

When you find an angle, check it against the con­di­tions given, such as $A \gt B$ or $A + B \le 90^\circ$. In iden­tity proofs, men­tion the restric­tion only when you can­cel a fac­tor that could be zero.

Is it accept­able to leave answers like 3/√10?#

Yes, unless the ques­tion asks for a ratio­nal denom­i­na­tor. If it does, write $\displaystyle \frac{3\sqrt{10}}{10}$.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
4. Loney, S. L. Plane Trigonom­e­try. Cam­bridge Uni­ver­sity Press.