Tanθ + tan(90-θ)=secθ + sec(90-θ)
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Sia ? 6 years, 1 month ago
We have to prove that: tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Here, LHS = tan θ + tan (90° – θ)
= tan θ + cot θ
=sinθcosθ+cosθsinθ
=sin2θ+cos2θsinθcosθ
=1sinθcosθ [∵ sin2 θ + cos2 θ = 1]
= cosec θ·sec θ
⇒ LHS = cosec θ·secθ
RHS = sec θ sec θ (90° - θ)
= sec θ cosec θ
⇒ LHS = RHS
Hence, verified.
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