NCERT Solutions Maths Class 10 Exercise 8.4

# NCERT Solutions Maths Class 10 Exercise 8.4

## Maths Class 10 Exercise 8.4

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution:  For sin A,

Let us use the identity, cosec2 A – cot2 A = 1

cosec2 A = 1 + cot2 A

For sec A,

Let us use the identity, sec2 A – tan2 A = 1

sec2 A = 1 + tan2 A

For tan A,

2. Write all the other trigonometric ratios of A in terms of sec A.

Solution: For sin A,

Let us use the identity, sin2 A + cos2 A = 1

sin2 A = 1 – cos2 A

For cos A,

For tan A,

Let us use the identity, sec2 A – tan2 A = 1

tan2 A = sec2 A – 1

For cot A,

For cosec A,

3. Evaluate:

(i)

(ii) sin 25° cos 65° + cos 25° sin 65°

Solution: (i)

=

=

[Since, sin (90° – Î¸) = cos Î¸ and cos (90° – Î¸) = sin Î¸]

= 1/1 = 1                   [Since, sin2 Î¸ + cos2 Î¸ = 1 ]

(ii) sin 25° cos 65° + cos 25° sin 65°

= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)

= sin 25° sin 25° + cos 25° cos 25°

[Since sin (90° – Î¸) = cos Î¸ and cos (90° – Î¸) = sin Î¸]

= sin2 25° + cos2 25° = 1                [ Since, sin2 Î¸ + cos2 Î¸ = 1]

4. Choose the correct option. Justify your choice.

(i) 9 sec2 A – 9 tan2 A =

(A) 1                         (B) 9                         (C) 8                       (D) 0

(ii) (1 + tan Î¸ + sec Î¸) (1 + cot Î¸  cosec Î¸) =

(A) 0                         (B) 1                         (C) 2                        (D) 1

(iii) (sec A + tan A) (1 sin A) =

(A) sec A                  (B) sin A                   (C) cosec A            (D) cos A

(iv) =

(A)  sec2 A                   (B)  –1                      (C) cot2 A                  (D) tan2 A

Solution: (i) (B) 9 sec2 A  – 9 tan2 A

= 9(sec2 A  – tan2 A)

= 9 × 1 = 9

(ii) (C) (1 + tan Î¸ + sec Î¸) (1 + cot Î¸ – cosec Î¸)

=

=

=

=

=           [Since, sin2Î¸ +cos2Î¸ =1]

=

= 2

(iii) (D) (sec A + tan A) (1 – sin A)

=

=

=     [Since, 1 – sin2 A = cos2 A]

(iv) (D)

=

= tan2 A

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

(i)

(ii)

(iii)

(iv)

(v), using the identity cosec2 A = 1 + cot2 A

(vi)

(vii)

(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

(ix) (cosec A – sin A) (sec A – cos A)

(x)

Solution: (i) L.H.S. = (cosec Î¸ – cot Î¸)2

= cosec2 Î¸ + cot2 Î¸ – 2 cosec Î¸ cot Î¸

=

=

=

=           [Since, a2 + b2 – 2ab = (a – b)2]

=

=

=

= R.H.S.

(ii) L.H.S. =

=

=

=                        [Since, sin2 Î¸ + cos2 Î¸ = 1]

=

= 2/cos A = 2 sec A = R.H.S.

(iii) L.H.S. =

=

=

=

=

=

=

[Since, a3 – b3 = (a –b) (a2 + b2 + ab)]

=     [Since, sin2 Î¸ + cos2 Î¸ = 1]

=

= 1 + sec Î¸ cosec Î¸

(iv) L.H.S. =

=

= 1 + cos A

=

=

=

= R.H.S.

(v) L.H.S. =

Dividing all the terms by sin A, we get

=

=

=

=

= cot A + cosec A

= R.H.S.

(vi) L.H.S. =

=

=              [Since, (a + b)(a – b) = a2 – b2]

=               [Since, 1 – sin2 Î¸ = cos2 Î¸]

=

= sec A + tan A

= R.H.S.

(vii) L.H.S. =

=

=    [Since, 1 – sinÎ¸ = cosÎ¸]

=

=

= tan Î¸

= R.H.S.

(viii) L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2

=

=

=

=

= 5 + cosec2 A + sec2 A

= 5 + 1 + cot2 A + 1 + tan2 A

[Since, cosec2 Î¸ = 1 + cot2 Î¸ and sec2 Î¸ = 1 + tan2 Î¸]

= 7 + tan2 A + cot2 A

= R.H.S.

(ix) L.H.S. = (cosec A – sin A) (sec A – cos A)

=

=

=

= sin A cos A

=                      [Since, sin2 Î¸ + cos2 Î¸ = 1]

Dividing the numerator and the denominator by sin A cos A, we get

=

=

=

= R.H.S.

(x) L.H.S. =

[Since, 1 + tan2 Î¸ = sec2 Î¸ and 1 + cot2 Î¸ = cosec2 Î¸]

=

= tan2 A = R.H.S.

Now, the middle term =

=

= (– tan A)2

= tan2

= R.H.S.