# Introduction to trignometry class 10 ncert solutions

## Important Formula (Trigonometry class ten) and Chapter Summary

In the right angle triangle ABC, the right angle is ∠ B.

The value of sin or cos never exceeds 1, but the value of sec and cosec is always greater than or equal to 1.

Exercise 8.1
Question – 1 - In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(a) Sin A, Cos A
Solution: AB = 24 cm, BC = 7 cm, AC = ?
The value of AC can be calculated by using Pythagoras Theorem:

(b) Sin C, Cos C
Solution:

Question – 2 - In the given figure, find tan P – cot R.

Solution: Value of QR can be calculated by using Pythagoras theorem:

Now;

Question – 3 - If sin A = ¾, calculate cos A and tan A.
Solution: Sin A = ¾ = p/h
We can calculate b by using Pythagoras theorem;

Now;

Question – 4 - Given 15 cot A = 8, find sin A and sec A.
Solution: 15 cot A = 8

This means, b = 8 and p = 15
We can calculate h by using Pythagoras theorem;

Now;

Question – 5 - Given sec θ = 13/12, calculate all other trigonometric ratios.
Solution:

This means, h = 13 and b = 12.
We can calculate p by using Pythagoras theorem;

Other trigonometric ratios can be calculated as follows:

Question – 6 - If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B
Solution: For no two different angles the cos ratio is same. (Ref: Table of trigonometric ratio).
Question – 7 - If cot θ = 7/8, evaluate:
Solution:

This means, b = 7 and p = 8.
We can calculate h by using Pythagoras theorem;

Solution:

Solution:

Question – 8 - If 3 cot A = 4, check whether  or not.
Solution: 3 cot A = 4 means cot A = 4/3 = b/p
Hence, p = 3 and b = 4.
We can calculate h by using Pythagoras theorem;

Now; the equation can be checked as follows:
LHS:

RHS:

It is clear that LHS = RHS.
Question – 9 - In triangle ABC, right-angled at B, if tan A =  find the value of:
Solution:

We can calculate h by using Pythagoras theorem;

(a) Sin A Cos C + Cos A Sin C
Solution:
Sin A Cos C + Cos A Sin C

(b) Cos A Cos C – Sin A Sin C
Solution:
Cos A Cos C – Sin A Sin C

Question – 10 - In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution: Given; PR + QR = 25 cm and PQ = 5 cm.
Hence, PR = 25 – QR
We can calculate PR and QR by using Pythagoras theorem;

Now;

Question – 11 - State whether the following are true or false. Justify your answer.
(a) The value of tan A is always less than 1.
Solution: False; value of tan begins from zero and goes on to become more than 1.
(b) sec A = 12/5 for some value of angle A.
Solution: True, value of cos is always more than 1.
(c) cos A is the abbreviation used for the cosecant of angle A.
Solution: False, cos is the abbreviation of cosine.
(d) cot A is the product of cot and A.
Solution: False, cot A means cotangent of angle A.
(e) sin A = 4/3 for some angle A.
Solution: False, value of sin is less than or equal to 1, while this value is more than 1.

## Solution of NCERT Exercise 8.2

Exercise 8.2
Question – 1 - Evaluate the following:
Solution:

Solution:

Solution:

Solution:

Solution:
Question – 2 - Choose the correct option and justify your choice:
Solution:

Solution:
(iii) Sin 2A = 2 Sin A is true when A = ?
1. 30°
2. 45°
3. 60°
Solution: Option (a) is correct answer.
Sin 0° = 0 and Sin 2 x 0° = Sin 0° = 0
Sin 30° = ½ But Sin 2 x 30° = Sin 60° is not equal to Sin 30°. The same holds true for sin 45°.
Solution:
Question – 3 - If , find A and B.
Solution: We know;
Hence, A + B = 60°
Similarly,
Hence, A – B = 30°
Adding equation (1) and (2), we get;
Question – 4 - State whether the following are true or false. Justify your answer.
(i) Sin (A+B) = Sin A + Sin B
Solution: False; this can be proved by assuming different values for A and B.
(ii) The value of sin θ increases as θ increases.
Solution: True, the value of sin θ increases from zero for zero degree and goes up to 1 for 90 degree.
(iii) The value of cos θ increases are θ increases.
Solution: False, the value of cos decreases from one to zero.
(iv) Sin θ = Cos θ for all values of θ.
Solution: False, values of sin and cos are equal only for 45 degree.
(v) Cot A is not defined for A = 0°
Solution: True, refer to values for trigonometric ratios.

## NCERT Solution ofExercise 8.3

Question – 1 - Evaluate
Solution:
Solution:
Solution:
Solution:
Question – 2 - Show that:
Solution:
Solution:
Question – 3 - If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution: tan 2A = cot (90° – 2A)
This means;
90° – 2A = A – 18°
Or, 108° – 2A = A
Or, 3A = 108°
Or, A = 108/3 = 36°
Question – 4 - If tan A = cot B, prove that A + B = 90°
Solution: tan A = cot (90°- A) = cot B
Or, A + B = 90° proved.
Question – 5 - If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution: sec 4A = cosec (90° - 4A) = cosec (A - 20°)
This means;
90° – 4A = A - 20°
Or, 110°– 4A = A
Or, 5A = 110°
Or, A = 22°
Question – 6 - If A, B and C are interior angles of a triangle ABC, then show that
Solution: Since, A + B + C = 90°
So, B + C = 90°- A
Hence,
Question – 7 - Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution: sin 67° + cos 75° can be written as follows:
Sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

## NCERT Solution ofExercise 8.4

Question – 1 - Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution: sin A can be expressed in terms of cot A, as follows:

sec A can be expressed in terms of cot A, as follows:

tan A can be expressed in terms of cot A, as follows:

Question – 2 - Write all the other trigonometric ratios of A in terms of sec A.
Solution: Sin A can be written in terms of sec A as follows:

cos A can be written in terms of sec A as follows:

tan A can be written in terms of sec A as follows:

cosec A can be written in terms of sec A as follows:

cot A can be written in terms of sec A as follows:

Question – 3 - Evaluate:

Solution:

Solution:

Question – 4 - Choose the correct option. Justify your choice.

Solution: (b) is the correct option. Following is the explanation:

Solution: (c) is the correct option. Following is the explanation:

Solution: (d) is the correct option. Following is the explanation:

Solution: (d) is the correct option. Following is the explanation:

Question – 5 - Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Solution:

LHS = RHS proved

Solution:

Solution:

Solution:

using the identity
Solution:

After dividing numerator and denominator by sin A, we get

After multiplying with [cotA – (1-cosecA)] in numerator and denominator both we get

Solution:

Solution:

Solution:

Hence; LHS = RHS proved

Solution:

Hence; LHS = RHS proved

Solution:

Hence; LHS = Middle Term = RHS proved