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Question
Question 12
Prove that 1+sec θ−tan θ1+sec θ+tan θ=1−sin θcos θ
Prove that 1+sec θ−tan θ1+sec θ+tan θ=1−sin θcos θ
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Solution
LHS=1+sec θ−tan θ1+sec θ+tan θ
=1+1cos θ−sin θcos θ1+1cos θ+sin θcos θ[∵sec θ=1cos θ and tan θ=sin θcos θ]
=cos θ+1−sin θcos θ+1+sin θ=(cos θ+1)−sin θ(cos θ+1)+sin θ=2 cos2θ2−2 sinθ2.cosθ22 cos2θ2+2 sin θ2.cos θ2
[∵1+cos θ=2 cos2 θ2 and sin θ=2 sin θ2.cos θ2]
=2 cos2 θ2−2 sin θ2.cos θ22 cos2 θ2+2 sin θ2.cos θ2=2 cos θ2(cos θ2−sin θ2)2 cos θ2(cos θ2+sin θ2)
=cos θ2−sin θ2cos θ2+sin θ2×(cos θ2−sin θ2)(cos θ2−sin θ2) [by rationalization]
=(cos θ2−sin θ2)2(cos2 θ2−sin2 θ2)
[∵(a−b)2−a2+b2−2ab and (a−b)(a+b)=(a2−b2)]
=(cos2 θ2+sin2 θ2)−(2 sin θ2.cos θ2)cos θ [∵cos2 θ2−sin2 θ2=cos θ]
=1−sin θcos θ [∵sin2 θ2+cos2 θ2=1]
=RHS
=1+1cos θ−sin θcos θ1+1cos θ+sin θcos θ[∵sec θ=1cos θ and tan θ=sin θcos θ]
=cos θ+1−sin θcos θ+1+sin θ=(cos θ+1)−sin θ(cos θ+1)+sin θ=2 cos2θ2−2 sinθ2.cosθ22 cos2θ2+2 sin θ2.cos θ2
[∵1+cos θ=2 cos2 θ2 and sin θ=2 sin θ2.cos θ2]
=2 cos2 θ2−2 sin θ2.cos θ22 cos2 θ2+2 sin θ2.cos θ2=2 cos θ2(cos θ2−sin θ2)2 cos θ2(cos θ2+sin θ2)
=cos θ2−sin θ2cos θ2+sin θ2×(cos θ2−sin θ2)(cos θ2−sin θ2) [by rationalization]
=(cos θ2−sin θ2)2(cos2 θ2−sin2 θ2)
[∵(a−b)2−a2+b2−2ab and (a−b)(a+b)=(a2−b2)]
=(cos2 θ2+sin2 θ2)−(2 sin θ2.cos θ2)cos θ [∵cos2 θ2−sin2 θ2=cos θ]
=1−sin θcos θ [∵sin2 θ2+cos2 θ2=1]
=RHS
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