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