Q. 54.2( 672 Votes )

# 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

(vi)

(vii)

(viii)

(ix) [Hint: Simplify LHS and RHS separately]

(x)

Answer :

**(i)**

To Prove:

Proof:

=

=

Since,

=

= RHS

Hence, proved

**(ii)**

To Prove:

Proof:

LHS = +

Use the identity sin^{2}θ + cos^{2}θ = 1

=

=

=

=

= 2 sec A

= RHS

**(iii)**

[Hint: Write the expression in terms of sin θ and Cos θ]

=

Use the formula a^{3} - b^{3} = (a^{2} + b^{2} + ab)(a - b)

Cancelling (sin θ - cos θ) from numerator and denominator

As cosθ = 1/secθ and sinθ = 1/cosecθ

= 1 + sec θ cosec θ

= RHS

**(iv)**

[Hint: Simplify LHS and RHS separately]

**LHS **

use the formula secA = 1/cosA

= cos A + 1

**RHS**

Use the identity sin^{2}θ + cos^{2}θ = 1

Use the formula a^{2} - b^{2} = (a - b) ( a + b )

= cos A + 1

LHS = RHS

**(v)** using the identity

**LHS**Dividing Numerator and Denominator by sin A

Use the formula cotθ = cosθ / sinθ

Using the identity

Use the formula a

^{2}- b

^{2}= (a - b) ( a + b )

= cot A + cosec A

= RHS

**(vi)**

Dividing numerator and denominator of LHS by cos A

As cosθ = 1/secθ and tanθ = sinθ /cosθ

Rationalize the square root to get,

Use the formula a^{2} - b^{2} = (a - b) ( a + b ) to get,

Use the identity sec^{2}θ = 1 + tan^{2}θ to get,

= sec A + tan A

= RHS

**(vii)**

To Prove:

Proof:**LHS**Since

As tanθ = sinθ/cosθ

= tanθ

= R.H.S

Hence, Proved.

**(viii)**

To Prove:

Proof:**LHS** = (sin A + cosec A)^{2} + (cos A + sec A)^{2}

Use the formula (a+b)^{2} = a^{2} + b^{2} + 2ab to get,

= (sin^{2}A + cosec^{2}A + 2sin A cosec A) + (cos^{2}A + sec^{2}A + 2 cos A sec A)

Since sinθ=1/cosecθ and cosθ = 1/secθ

= sin^{2}A + cosec^{2}A + 2 + cos^{2}A + sec^{2}A + 2

=(sin^{2}A + cos^{2}A)+cosec^{2}A+sec^{2}A+2+2

Use the identities sin^{2}A + cos^{2}A = 1, sec^{2}A = 1 + tan^{2}A and cosec^{2}A = 1 + cot^{2}A to get

= 1+ 1 + tan^{2}A + 1 + cot^{2}A + 2 + 2

= 1 + 2 + 2 + 2 + tan^{2}A + cot^{2}A

= 7 + tan^{2}A + cot^{2}A

= RHS

**(ix)**

To Prove:

[Hint: Simplify LHS and RHS separately]

Proof:**LHS** = (cosec A – sin A) (sec A – cos A)

Use the formula sinθ = 1/cosecθ and cosθ = 1/secθ

= cos A sin A

**RHS**

use the formula tanθ=sinθ/cosθ and cotθ=cosθ/sinθ

Use the identity sin

^{2}A + cos

^{2}A = 1

= cos A sin A

LHS = RHS

**(x)**

To Prove:

Proof:

Taking left most term

Since, cot A is the reciprocal of tan A, we have

= right most part

Taking middle part:

= right most part

Hence, Proved.

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