# Chapter 8 Introduction to Trigonometry and its Equation NCERT Exemplar Solutions Exercise 8.3 Class 10 Maths

 Chapter Name NCERT Maths Exemplar Solutions for Chapter 8 Introduction to Trigonometry and its Equation Exercise 8.3 Book Name NCERT Exemplar for Class 10 Maths Other Exercises Exercise 8.1Exercise 8.2Exercise 8.4 Related Study NCERT Solutions for Class 10 Maths

### Exercise 8.3 Solutions

Prove the following (from Q.1  to Q.7):

1. SinÎ¸/(1 + cosÎ¸) + (1 + cosÎ¸)/sinÎ¸ = 2cosecÎ¸

Solution

L.H.S =

Hence proved.

2. tanA/(1 + secA) - tanA/(1 - secA) = 2cosecA

Solution

L.H.S

As,
sec2A – tan2A = 1
⇒ sec2A – 1  = tan2A
= R.H.S
Hence proved.

3. If tan A = 3/4, then sin A cos A = 12/25

Solution

As given in the question,
tan A = 3/4
Also,
tan A = Perpendicular/Base
So,
tan A = 3k/4k
Where,
Perpendicular = 3k
Base = 4k

Using Pythagoras Theorem,
(hypotenuse)2 = (perpendicular)2 + (base)2
⇒ (hypotenuse)2 = (3k)2 + (4k)2
= 9k2 + 16k2
= 25k2
hypotenuse = 5k
Now find sin A and cos A,
sin A = 3k/5k
= 3/5
cos A = 4k/5k
= 4/5
So,
sin A cos A = 3/5 × 4/5
= 12/25
Hence, proved.

4. (sinÎ± + cosÎ±)(tanÎ± + cotÎ±) = secÎ± + cosecÎ±

Solution

L.H.S:
(sinÎ± + cosÎ±)
So,
(sinÎ± + cosÎ±)(tanÎ± + cotÎ±)

secÎ± + cosecÎ±
= R.H.S
Hence, proved.

5. (√3 + 1)(3 - cot30°) = tan3 60° - 2sin60°

Solution

(√3 + 1)(3 - cot30°)
= (√3 + 1)(3 - √3)
[as, cot 30° = √3]
= (√3 + 1) √3(√3 - 1)
[ as, (3 - √3) = √3(√3 - 1)]
= [(√3)2 - 1]√3
[as, (√3 + 1)(√3 - 1) = [(√3)2 - 1]
= (3 - 1)√3
= 2√3
Also, solving
R.H.S :
tan3 60° - 2sin 60°
As,
tan 60° = √3 and
sin 60° = √3/2,
we get,
(√3)3 - 2(√3/2) = 3√3 - √3 = 2√3
So,
L.H.S = R.H.S
Hence, proved.

6. 1 + cot2 Î±/(1 + cosec Î±) = cosec Î±

Solution

Taking LHS,

7. tan Î¸ + tan (90° - Î¸) = sec Î¸ sec(90 - Î¸)
Solution
L.H.S = tan (90° - Î¸) = cot Î¸
tan Î¸ + tan (90 - Î¸) = tan Î¸ + cot(Î¸)

= sec Î¸ cosec Î¸
= sec Î¸ sec(90 - Î¸)
Hence proved.

8. Find the angle of elevation of the sun when the shadow of a pole h metres high is √3h metres long.
Solution
Let the angle of elevation of the Sun is Î¸.
Given, height of pole = h m
Now, in ∆ABC,

tan Î¸ = 1/√3
= tan 30°
Î¸ = 30°
So, the angle of elevation of the Sun is 30°.

9. If √3 tan Î¸ = 1, then find the value of sin2Î¸ - cos2Î¸.
Solution
Given,
√3 tan Î¸ = 1,
⇒ tanÎ¸ = 1/√3
⇒ tanÎ¸ = tan 30°
Î¸ = tan 30°
So,
sin2 Î¸ - cos2 Î¸ = sin2 30° - cos2 30°
= (1/2)2 - (√3/2)2
= -1/2

10. A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.
Solution
Given,
height of the ladder = 15 m
Let the height of the vertical wall = h
Because, and the ladder makes an angle of 60° with the wall i.e., Î¸ = 60°
In ∆QPR,

cos60° = PR/PQ
1/2 = h/15
h = 15/2 m

11. Simplify (1 + tan2 Î¸)(1 - sinÎ¸)(1 + sinÎ¸)
Solution
(1 + tan2 Î¸)(1 - sinÎ¸)(1 + sinÎ¸)
= (1 + tan2 Î¸)(1 - sin2 Î¸)  [as, (a - b)(a + b) = a2 - b2]
= sec2 Î¸.cos2 Î¸  [as, 1 + tan2 Î¸= sec2 Î¸ and cos2 Î¸ + sin2 Î¸ = 1]
= (1/cos2 Î¸).cos2 Î¸
= 1  [as, sec Î¸ = 1/cosÎ¸]

12. If 2 sin2 Î¸ - cos2 Î¸ = 2, then find the value of Î¸.
Solution
Given, 2 sin2 Î¸ - cos2 Î¸ = 2
2 sin2 Î¸ - (1 - sin2 Î¸) = 2  [as, sin2 Î¸ + cos2 Î¸ = 1]
⇒ 2sin2 Î¸ - sin2 Î¸ - 1 = 2
⇒ 3sin2 Î¸ = 3
⇒ sin2 Î¸ = 1
⇒ sinÎ¸ = 1 = sin90°  [as, sin 90° = 1]
⇒ Î¸ = 90°

13. Show that [cos2 (45° + Î¸) + cos2 (45° - Î¸)/tan(60° + Î¸)tan(30° - Î¸) = 1
Solution

14. An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution
Let the angle of elevation of the top of the tower from the eye of the observer is Î¸
Given that,

AB = 22m,
PQ = 1.5 m = MB
and
QB = PM = 20.5 m
Also,
AM = AB - MB
22 - 1.5 = 20.5 m
Now in ∆APM,
tanÎ¸ = AM/PM
=20.5/20.5
=1
= tan45°
tanÎ¸ = tan45°
⇒ Î¸ = 45°
Hence, the required angle of elevation of the top of the tower from the eye of the observer is
45°.

15. Show that tan4 Î¸ + tan2 Î¸ = sec4 Î¸ - sec2 Î¸.
Solution
LHS = tan4 Î¸ + tan2 Î¸
= tan2 Î¸(tan2 Î¸ + 1)
= tan2 Î¸.sec2 Î¸  [as, sec2 Î¸ = tan2 Î¸ + 1]
= (sec2 Î¸ - 1).sec2 Î¸  [as, tan2 Î¸ = sec2 Î¸ - 1]
= sec4 Î¸ - sec2 Î¸
= RHS