Listen NCERT Audio Books - *Kitabein Ab Bolengi*

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

(vi)

(vii)

(viii)

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

(x)

(i)

LHS = (cosec θ - cot θ)

= ( – )^{2}

=

= (1 - cos θ)^{2}/sin^{2} θ

= (1 – cos θ)^{2}/1 – cos^{2} θ

= (1 – c0s θ)^{2}/ (1 – cos θ) (1 + cos θ)

=

= RHS

(ii)

LHS = +

= cos^{2}A + (1 + sin^{2}A)/(1 + sinA) (cos A)

= cos^{2}A + 1 + sin^{2}A + 2sinA/(1 + sinA) (cosA)

= sin^{2}A + cos^{2}A + 1 + 2sinA/(1 + sinA) (cosA)

=

=

=

=

= 2 sec A

= RHS

(iii)

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

=

=

= sin^{2} θ/[cos θ (sin θ – cos θ)] + cos^{2} θ/ [sin θ (cos θ – sin θ)]

= sin^{2} θ/[cos θ (sin θ - cos θ)] – cos^{2} θ/[sin θ (sin θ - cos θ)]

= 1/(sin θ - cos θ) [(sin^{2} θ/cos θ) – (cos^{2} θ/sin θ)]

= 1/(sin θ – cos θ) * [(sin ^{3} θ – cos^{3} θ)/sin θ cos θ]

= [(sin θ – cos θ) (sin^{2} θ + cos^{2} θ + sin θcos θ)]/ [sin θ - cos θ) sin θ cos θ]

= (1 + sin θ cos θ)/sin θ cos θ

= 1/sin θ cos θ + 1

= 1 + sec θ cosec θ

= RHS

(iv)

[Hint: Simplify LHS and RHS separately]

LHS = (1 + secA)/ secA

= (1 + 1/cos A)/1/cos A

= (cos A + 1)/cosA/1/cosA

= cos A + 1

RHS = sin^{2}A/(1 – cos A)

= (1 – cos^{2}A)/(1 – cos A)

= (1 – cos A) (1 + cos A)/ (1 – cos A)

= cos A + 1

LHS = RHS

(v) using the identity

LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)

Dividing Numerator and Denominator by sin A

= (c0s A – sin A + 1) / sin A/ (cos A sin A – 1) / sin A

= (cot A – 1 + cosec A)/ (cot A + 1 – cosec A)

= (cot A – cosec^{2}A + cot^{2}A + cosec A)/ (cot A + 1- cosec A) [Using cosec^{2}A – cot^{2}A = 1]

= [(cot A + cosec A) – (cosec A + cot A) (cosec A – cot A)]/ (1 – cosec A + cot A)

= (cot A + cosec A) (1 – cosec A + cot A)/ (1 – cosec A + cot A)

= cot A + cosec A

= RHS

(vi)

Dividing numerator and denominator of LHS by cos A

=

=

= *

= [(sec A + tan A)^{2}/sec^{2}A – tan^{2}A]^{1/2}

=

= sec A + tan A

= RHS

(vii)

(sin θ – 2 sin^{3} θ)/ (2cos^{3} θ – cos θ)

= [sin θ (1 – 2sin^{2} θ)]/[cos θ (2cos^{2} θ – 1)]

= sin θ [1 – 2 (1 – cos^{2} θ)] / [cos θ (2 cos^{2} θ – 1)]

= [sin θ (2cos^{2} θ – 1)] / [cos θ (2cos^{2} θ – 1)]

= tan θ = RHS

(viii)

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

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

= (sin^{2}A + cos^{2}A) + 2 sin A (1/sinA) + 2cos A (1/cosA) + 1 + tan^{2}A + 1 + cot^{2}A

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

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

= RHS

(ix)

[Hint: Simplify LHS and RHS separately]

LHS = (cosec A – sin A) (sec A – cos A)

= (1/sin A – sin A) (1/cos A – cos A)

= [(1 – sin^{2}A)/ sinA] [(1 – cos^{2}A)/cosA]

= (cos^{2}A/sin A) * (sin^{2}A/cos A)

= cos A sin A

RHS = 1/(tan A + cot A)

= 1 / (sin A/cos A + cos A / sin A)

= 1/[(sin^{2}A + cos^{2}A)/sin A cos A]

= cos A sin A

LHS = RHS

(x)

LHS = (1 + tan^{2}A/ 1 + cot^{2}A)

= (1 + tan^{2}A / 1 + 1/tan^{2}A)

= 1 + tan^{2}A/ [(1 + tan^{2}A)/tan^{2}A]

= tan^{2}A

6