Question

cos2a/1-sin2a=tan (45+a)

Answer 1

we have to prove that cos2A/(1 - sin2A) = tan(45° + A)

we know, sin2x = 2tanx/(1 + tan²x)

cos2x = (1 - tan²x)/(1 + tan²x)

LHS = cos2A/(1 - sin2A)

= {(1 - tan²A)/(1 + tan²A)}/{1 - 2tanA/(1 + tan²A)}

= (1 - tan²A)/(1 + tan²A - 2tanA)

= {(1 - tanA)(1 + tanA)}/(1 - tanA)²

= (1 + tanA)/(1 - tanA)

[ we know tan45° = 1, putting 1 = tan45° ]

= (tan45° + tanA)/(1 - tan45°.tanA)

now using formula,

tan(C + D) = (tanC + tanD)/(1 - tanC.tanD)

so, (tan45° + tanA)/(1 - tan45°.tanA) = tan(45° + A) = RHS

here LHS = RHS

hence proved.

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Answer 2

Answer:

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