NCERT Solutions for Class 10 Maths Ex 8.3

# NCERT Solutions for Class 10 Maths Ex 8.3

## NCERT Solutions for Class 10 Maths Ex 8.3

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, cosecA – cotA = 1

cosecA = 1 + cotA

For sec A,

Let us use the identity, secA – tanA = 1

secA = 1 + tanA

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, sinA + cosA = 1

sinA = 1 – cosA

For cos A,

For tan A,

Let us use the identity, secA – tanA = 1

⇒ tanA = secA – 1

For cot A,

For cosec A,

3. 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 – sinA = cos2 A]

(iv) (D)

=

= tan2 A

4. 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 cosec2A = 1 + cot2A.

(vi)

(vii)

(viii) (sin A + cosec A)2 + (cos A + sec A)7 + tanA + cotA

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

(x)

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

= cosecÎ¸ + cotÎ¸ – 2 cosec Î¸ cot Î¸

=

=

=

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

=

=

=

= R.H.S.

(ii) L.H.S. =

=

=

=                        [Since, sinÎ¸ + cosÎ¸ = 1]

=

= 2/cos A = 2 sec A = R.H.S.

(iii) L.H.S. =

=

=

=

=

=

=

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

=     [Since, sinÎ¸ + cosÎ¸ = 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 – sinÎ¸ = cosÎ¸]

=

= 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 + cosecA + secA

= 5 + 1 + cotA + 1 + tanA

[Since, cosecÎ¸ = 1 + cotÎ¸ and secÎ¸ = 1 + tanÎ¸]

= 7 + tanA + cotA

= R.H.S.

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

=

=

=

= sin A cos A

=                      [Since, sinÎ¸ + cosÎ¸ = 1]

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

=

=

=

= R.H.S.

(x) L.H.S. =

[Since, 1 + tanÎ¸ = secÎ¸ and 1 + cotÎ¸ = cosecÎ¸]

=

= tanA = R.H.S.

Now, the middle term =

=

= (– tan A)2

= tan

= R.H.S.