##### 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/sin2 θ

= (1 – cos θ)2/1 – cos2 θ

= (1 – c0s θ)2/ (1 – cos θ) (1 + cos θ)

=

= RHS

(ii)

LHS = +

= cos2A + (1 + sin2A)/(1 + sinA) (cos A)

= cos2A + 1 + sin2A + 2sinA/(1 + sinA) (cosA)

= sin2A + cos2A + 1 + 2sinA/(1 + sinA) (cosA)

=

=

=

=

= 2 sec A

= RHS

(iii)

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

=

=

= sin2 θ/[cos θ (sin θ – cos θ)] + cos2 θ/ [sin θ (cos θ – sin θ)]

= sin2 θ/[cos θ (sin θ - cos θ)] – cos2 θ/[sin θ (sin θ - cos θ)]

= 1/(sin θ - cos θ) [(sin2 θ/cos θ) – (cos2 θ/sin θ)]

= 1/(sin θ – cos θ) * [(sin 3 θ – cos3 θ)/sin θ cos θ]

= [(sin θ – cos θ) (sin2 θ + cos2 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 = sin2A/(1 – cos A)

= (1 – cos2A)/(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 – cosec2A + cot2A + cosec A)/ (cot A + 1- cosec A) [Using cosec2A – cot2A = 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/sec2A – tan2A]1/2

=

= sec A + tan A

= RHS

(vii)

(sin θ – 2 sin3 θ)/ (2cos3 θ – cos θ)

= [sin θ (1 – 2sin2 θ)]/[cos θ (2cos2 θ – 1)]

= sin θ [1 – 2 (1 – cos2 θ)] / [cos θ (2 cos2 θ – 1)]

= [sin θ (2cos2 θ – 1)] / [cos θ (2cos2 θ – 1)]

= tan θ = RHS

(viii)

LHS = (sin A + cosec A)2 + (cos A + sec A)2

= (sin2A + cosec2A + 2sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)

= (sin2A + cos2A) + 2 sin A (1/sinA) + 2cos A (1/cosA) + 1 + tan2A + 1 + cot2A

= 1 + 2 + 2 + 2 + tan2A + cot2A

= 7 + tan2A + cot2A

= 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 – sin2A)/ sinA] [(1 – cos2A)/cosA]

= (cos2A/sin A) * (sin2A/cos A)

= cos A sin A

RHS = 1/(tan A + cot A)

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

= 1/[(sin2A + cos2A)/sin A cos A]

= cos A sin A

LHS = RHS

(x)

LHS = (1 + tan2A/ 1 + cot2A)

= (1 + tan2A / 1 + 1/tan2A)

= 1 + tan2A/ [(1 + tan2A)/tan2A]

= tan2A

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