Soln:
L.H.S.
= [1 + Tan2(p/4-A)]/[1 – Tan2(p/4-A)]
= 1/[{1 – Tan2(p/4-A)}/{ 1 + Tan2(p/4-A)}]
= 1/[cos2(p/4-A)] [Since, cos2A = (1 – Tan2A)/(1 + Tan2A)]
= 1/[cos(p/2-2A)]
= 1/sin2A
= cosec2A = R.H.S
Q.Prove that: [1+Tan^2(pie/4-A)] / [1-Tan^2(pie/4-A)] = cosec2A
![Q.Prove that: [1+Tan^2(pie/4-A)] / [1-Tan^2(pie/4-A)] = cosec2A](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEinGiOEaD5LPLqioiclipXvWEh8E5qSkBGwjNqIAtw0LjlC5ZpiCgwAGGCj53_TiIRRs6-n_Uq_8iH-vDjhMt9xytukA2QzZaedzC0_hwpVoO2qe45Cn6Dmm0oIYOz_dkS5aA-ee6BnzIim/s72-c/coollogo_com-4440765.png)
Q.Prove that: [1+Tan^2(pie/4-A)] / [1-Tan^2(pie/4-A)] = cosec2A
S.E.E. Solution
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Soln: L.H.S. = [1 + Tan 2 ( p /4-A)]/[1 – Tan 2 ( p /4-A)] = 1/[{1 – Tan 2 ( p /4-A)}/{ 1 + Tan 2 ( p /4-A)}] = 1/[cos2( p /4-A)...
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