NCERT Solutions for Class 10 Maths Chapter 10 Introduction to Trigonometry

$$1.\space \frac{1-tan^2 45^\circ}{1+tan^2 45^\circ} \space is \space equal \space to \space:$$

(a) tan 90°

(c) sin 45°

(b) 1

(d) sin 0°

• Ans. (d) sin 0°
Explanation :

Given,

$$\frac{1-tan^2 45^\circ}{1+tan^2 45^\circ} \space = \space \frac{1-1}{1+1} \space = \space 0 \space = \space sin \space 0^\circ .$$

Q. (1+tan2A)/(1+cot2A) equals to:
(a) sec2 A
(c) cot2 A

(b) – 1

(d) tan2 A
Ans. (d) tan2 A
Explanation :

Given,

$$\frac{1-tan^2 A}{1+tan^2 A} \space = \space \frac{1+ \frac{sin^2 A}{cos^2 A}}{1+ \frac{cos^2 A}{sin^2 A}} =\space \frac{ \frac{cos^2 A + sin^2 A}{cos^2 A}}{\frac{sin^2 A + cos^2 A}{sin^2 A}}$$
$$= \space \frac{(cos^2 A + \space sin^2 A) \sin^2 A}{(sin^2 A + \space cos^2 A) \cos^2 A}$$
$$= \space \frac{sin^2 A}{cos^2 A} \space \space \space \space (\because \space sin^2\space A\space +\space cos^2 A\space = \space 1)$$
$$=tan^2 A.$$

Q. If cos A =4/3, then the value of tan A is:

(a) 3/5

(c) 4/3

(b) 3/4

(b) 5/3

Ans. (b) 3/4

Explanation :

$$Given, \space cosA \space = \frac{4}{5}$$
$$\therefore \space \space \space \space sin A\space = \sqrt{1-cos^2\space }\begin{bmatrix} \because sin^2A + cos^2 A = 1 \\ \therefore sin A = \sqrt{1-cos^2 A} \end{bmatrix}\space$$
$$=\sqrt{1-\left(\frac{4}{5}\right)^2} \space = \sqrt{1-\frac{16}{25}} \space = \space \sqrt{\frac{9}{25}} \space = \space \frac{3}{5}$$
$$Now,\space tan A = \frac{sinA}{cosA} = \frac{\frac{3}{5}}{\frac{4}{5}} \space = \frac{3}{4} .$$

Q. Find the value of 9 sec2 A – 9 tan2 A.

Sol. Given,

9 sec2 A – 9 tan2 A = 9(sec2 A – tan2 A)

= 9(1) [... sec2 q – tan2 q = 1]

= 9. Ans.

Q. Find the value of (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Sol. Given,

• (1 + tan θ + sec θ ) (1 + cot θ – cosec θ )
• = [1 + tan θ + sec θ + cot θ + tan θ cot θ + sec θ cot θ – cosec θ – tan θ cosec θ – sec θ cosec θ ]
• = [1 + tan θ + sec θ + cot θ + 1 + cosec θ – cosec θ – sec θ – sec θ cosec θ ]
• = [2 + tan θ + cot θ – sec θ cosec θ ]
• = 2 + sin θ/Cos θ + cos θ/sin θ - 1/sin θ cos θ
• = 2 + ((sin 2 θ + cos 2 θ)-1)/sin θ cos θ
• = 2+((1-1)/sin θ cos θ) [... sin2 θ + cos2 θ = 1]
• = 2

Q. Find (sec A + tan A) (1 – sin A).

Sol. Given,

$$(secA + tanA) \space (1- sinA)$$
$$=\space \left(\frac{1}{cos A} + \space \frac{sin A}{cos A}\right) (1-sin\space A)$$
$$= \left(\frac{1+sinA}{cos A}\right)(1-sinA)$$
$$= \left(\frac{1-sin^2 A}{cos A}\right)$$
$$= \frac{cos^2 A}{cos A} \space \space [\because \space 1-sin^2 A = cos^2A]$$
$$= cos A. \space \space \space \space \space \space \space Ans.$$

Q. In ∆OPQ right angled at P, OP = 7 cm , OQ – PQ = 1 cm. Determine the values of sin Q and cos Q.

Sol. In ∆OPQ, we have

• OQ2 = OP2 + PQ2
• ⇒ (PQ + 1)2 = OP2 + PQ2
• [·.· OQ – PQ = 1 ⇒ OQ = 1 + PQ]
• ⇒ PQ2 + 2PQ + 1 = OP2 + PQ2
• ⇒ 2PQ + 1 = 72
• ⇒ 2PQ + 1 = 49
• ⇒ 2PQ = 48
• ⇒ PQ = 24 cm
• and OQ – PQ = 1 cm
• OQ = (PQ + 1) cm
• = 25 cm
• Thus, sin Q =QP/OQ= 7/25
• and cos Q =PQ/OQ=24/25 Ans.

Q. If ∠B and ∠Q are acute angles such that sin B = sin Q, then prove that ∠B = ∠Q.

• Sol. Consider two right angled triangles ABC and PQR such that sin B = sin Q.
• We have, sin B = AC/AB and sin Q =PR/PQ

$$\because \space \space \space \space \space sinB\space = \space sin\space Q$$
$$\Rightarrow \space \space \space \space \frac{AC}{AB} = \frac{PR}{PQ} \space \space \space \space (\because sin \space B\space = \space sin \space Q)$$
$$\Rightarrow \space \space \space \space \frac{AC}{PR} \space = \space \frac{AB}{PQ} \space = \space k \space (say), \space \space \space \space \space \space \space …(i)$$
$$\Rightarrow \space \space \space \space \space AC\space =\space k \space PR\space and\space AB\space =\space k\space PQ$$

In △ABC and △PQR by Pythagoras theorem, we
have
$$AB^2 \space = \space AC^2 + \space BC^2$$
$$and \space \space \space \space PQ^2 \space = \space PR^2 + QR^2$$
$$\Rightarrow \space \space \space \space \space \space\space BC = \sqrt{AB^2 \space – \space AC^2}$$
$$and \space \space \space \space \space \space QR \space = \space \sqrt{{PQ^2-PR^2}}$$
$$\Rightarrow \space \space \space \space \space \space \frac{BC}{QR}\space =\space \frac{\sqrt{AB^2-AC^2}}{\sqrt{PQ^2 – PR^2}}$$
$$=\frac{\sqrt{k^2PQ^2-k^2PR^2}}{\sqrt{PQ^2-PR^2}}$$
$$\Rightarrow \space \space \space \space \frac{BC}{QR} \space =\space \frac{k\sqrt{PQ^2-PR^2}}{\sqrt{PQ^2-PR^2}} = k \space \space \space \space \space …(ii)$$
From (i) and (ii), we have
$$\frac{AC}{PR} = \frac{AB}{PQ} = \frac{BC}{QR}$$
$$\Rightarrow \space \space \space \space \space \space \Delta ACB\sim \Delta PRQ$$
$$\Rightarrow \space \space \space \space \space \space \angle B = \angle Q. \space \space \space \space \space \space \space Hence\space Proved.$$

Q. In ∆ABC right – angled at C, AB = 29 units, BC = 21 units and ∠ABC = θ. Determine the values of (i) cos2 θ + sin2 θ, and (ii) cos2 θ – sin2 θ.
• Ans. (i) 1, (ii)41/841

Q. In ∆OPQ, right-angled at P, OP = 7 cm, OQ – PQ = 1 cm. Determine the values of sin Q and
cos Q.

• Ans. sin Q =7/25, cos Q = 24/25

Q. In ∆PQR, right-angled at Q, PQ = 5 cm, PR + QR = 25 cm. Determine the values of sin P, cos P and tan P.

• Ans. sin P =12/13, cos P =5/13 and tan P = 12/5.

Q. In ∆ABC, right-angled at B, if tan A =1/√3, find the value of (i) sin A cos C + cos A sin C, and (ii) cos A cos C – sin A sin C.

• Ans. (i) sin A cos C + cos A sin C = 1
• (ii) cos A cos C – sin A cos C = 0.

Q. In a ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine (i) sin A, cos A and (ii) sin C, cos C.

• Ans. (i) sin A = 7/25 and cos A = 24/25
• (ii) sin C = 24/25 and cos C = 7/25

Q. Given 15 cot A = 8. Find sin A and sec A.

• Ans. sin A = 15/17, sec A = 17/8·
Q. If cot θ = 7/8 evaluate (i) (1+sin θ)(1-sin θ)/(1+cos θ)(1-cos θ) and (ii) cot2 θ.
• Ans. (i) 49/64, (ii) 49/64·
Q. If 3 cot A = 4, check whether (1-tan2A)/(1+tan2A) = cos2 A – sin2 A or not.
• Ans. Both are equal to 7/25·

Q. If tan (A – B) = 1/√3 and tan (A + B) = √3, 0° < A + B ≤ 90° and A > B, find A and B.

• Ans. A = 45° and B = 15°.

Q. If sin (A – B) = 1/2 and cos (A + B) = 1/2, 0° < A + B ≤ 90° and A > B, find A and B.

• Ans. A = 45° and B = 15°.

Q. In ∆PQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.

• Ans. ∠P = 60° and ∠R = 30°.