Free download NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1 introduction to trigonometry in English Medium and Hindi Medium PDF form to use it offline. NCERT Textbook solutions for class 10 other subjects are also given to free use. Download Exercise 8.2 or Exercise 8.3 or Exercise 8.4 also in form or go for online use only.

## NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1

If you need solutions in Hindi, Click for Hindi Medium solutions of 10 Maths Exercise 8.1

### Class 10 Maths Exercise 8.1 Solutions in Hindi Medium

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##### Introduction to Trigonometry – Class 10 Maths

###### Practice Question Selected form CBSE Board Papers

- Question 1: If tan A = √2 – 1, show that sin A cos A = √2/4.
- Question 2: In a rectangle ABCD, AB = 20 cm and angle BAC = 60⁰. Find the measure of side BC and diagonals AC and BD. [Answer: BC = 20 √3 cm, AC = 40 cm, BD = 40 cm]
- Question 3: If A is an acute angle and cosec A = √5,
- (i) evaluate cot A – cosec A [Answer: (i) 2 – √5]
- (ii) Verify the identity sin²A + cos²A = 1.

- Question 4: If tan A + cot A = 2, find the value of tan²A + cot²A. [Answer: 2]
- Question 5: In ΔABC, right angled at angle B, AC – BC = 8 cm and AB = 12 cm. Find the value of sec C + 4 cot A. [Answer: 61/5]
- Question 6: If sec A + tan A = p, then find the value of cosec A. [Answer: [p² +1]/[p² – 1]]
- Question 7: If sin A = c/[√(c² + d²] and d > 0, find the value of cos A and tan A. [Answer: cos A = 2/[√(c² + d²)]; tan A = c/d]
- Question 8: In each of the following one of the six trigonometric ratio is given. Find the value of the other trigonometric ratios.
- (i) sin A = 7/25
- (ii) cos A = 9/41
- (iii) tan A = 8/15
- (iv) cosec A = √2
- (v) sec A = 13/12
- (vi) cot A = 20/21

[Answer: (i) cos A = 24/25 , tan A = 7/24, cosec A = 25/7 , sec A = 25/24 , cot A = 24/7

(ii) sin A = 40/41, tan A = 40/9, cosec A = 41/40 , sec A = 41/9 , cot A = 9/40

(iii) sin A = 8/17, cos A = 15/17, cot A = 15/8, cosec A = 17/8, sec A = 17/15

(iv) cos A = 1/√2, tan A = 1, sec A= √2, cot A = 1, sin A = 1/√2

(v) sin A = 5/13, cos A = 12/13, tan A = 5/12, cosec A 13/5, cot A = 12/5

(vi) sin A = 21/29, cos A = 20/29, tan A = 21/20, cosec A = 29/21, sec A = 29/20]