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(tan A)/(sec(A) - 1) + (tan A)/(sec(A) + 1) = 2cos ecA
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(tan A)/(sec(A) - 1) + (tan A)/(sec(A) + 1) = 2cos ecA

User Gwhyyy
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1 Answer

6 votes

Answer:

Explanation:

LHS = tanθ/secθ - 1 + tanθ/secθ + 1

Cross multiplication,

tanθ(1 - secθ) - tanθ(1 + secθ)/1 + sec^2θ

= tanθ(1 - secθ - 1 - secθ)/-tan^2θ

= -2secθ/-tanθ

= 2 x 1/cosθ / sinθ/cosθ

= 2cosecθ = RHS

Hence proved.

User Robert Co
by
8.2k points
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