Question

Tan(45+A/2)=secA+TanA

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Taking the LHS of the given equation, we have

$tan(45+\frac{A}{2})=\frac{tan45+tan\frac{A}{2}}{1-tan45tan\frac{A}{2}}$

=$\frac{1+tan\frac{A}{2}}{1-tan\frac{A}{2}}$

=$\frac{1+\frac{sin\frac{A}{2}}{cos\frac{A}{2}}}{1-\frac{sin\frac{A}{2}}{cos\frac{A}{2}}}}$

=$\frac{cos\frac{A}{2}+sin\frac{A}{2}}{cos\frac{A}{2}-sin\frac{A}{2}}$

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

=$\frac{cos^{2}\frac{A}{2}+sin^{2}\frac{A}{2}+2cos\frac{A}{2}sin\frac{A}{2}}{cosA}$

=$\frac{1+sinA}{cosA}$

=$\frac{1}{cosA}+\frac{sinA}{cosA}$

=$secA+tanA$

=RHS

Hence proved