Question #f8601

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Question #f8601

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Answer 1 · 2016-05-03T11:33:34

RHS=LHS proves the identity

Explanation:

To prove the identity, it may be simpler to start with RHS.

RHS= $\frac{tan (\frac{π}{4}) + tan (\frac{A}{2})}{1 - tan (\frac{π}{4}) tan (\frac{A}{2})}$ $(tan(pi/4)+tan(A/2))/(1-tan(pi/4)tan(A/2))$ = $\frac{1 + tan (\frac{A}{2})}{1 - tan (\frac{A}{2})}$ $(1+tan(A/2))/(1-tan(A/2)$

= $\frac{cos (\frac{A}{2}) + sin (\frac{A}{2})}{cos (\frac{A}{2}) - sin (\frac{A}{2})}$ $(cos(A/2)+sin (A/2))/(cos (A/2)- sin(A/2))$ = $\frac{cos (\frac{A}{2}) + sin (\frac{A}{2})}{cos (\frac{A}{2}) - sin (\frac{A}{2})} \cdot \frac{cos (\frac{A}{2}) + sin (\frac{A}{2})}{cos (\frac{A}{2}) + sin (\frac{A}{2})}$ $(cos(A/2)+sin (A/2))/(cos (A/2)- sin(A/2))* (cos(A/2)+sin (A/2))/(cos (A/2)+ sin(A/2))$

= $\frac{{cos}^{2} (\frac{A}{2}) + {sin}^{2} (\frac{A}{2}) + 2 sin (\frac{A}{2}) cos (\frac{A}{2})}{{cos}^{2} (\frac{A}{2}) - {sin}^{2} (\frac{A}{2})}$ $(cos^2 (A/2) +sin^2 (A/2) +2sin(A/2) cos( A/2))/(cos^2 (A/2) - sin^2 (A/2))$

$= \frac{1 + sin A}{cos A} = sec A + tan A = L H S$ $=(1+sinA)/cosA = secA +tan A= LHS$

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Question #f8601

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