Question

Prove : tan^2A/(secA-1)^2=1+cosA/1-cosA

Answer 1

ye raha is question ka solution.....

Answer 2

Answer:

$\frac{tan^{2}A}{(secA-1)^{2}}=\frac{1+cosA}{1-cosA}$

Step-by-step explanation:

$LHS=\frac{tan^{2}A}{(secA-1)^{2}}$

$=\frac{(sec^{2}A-1)}{(secA-1)^{2}}\\=\frac{(secA+1)(secA-1)}{(secA-1)(secA-1)}\\=\frac{(secA+1)}{(secA-1)}\\=\frac{\frac{1}{cosA}+1}{\frac{1}{cosA}-1}\\=\frac{\frac{(1+cosA)}{cosA}}{\frac{(1-cosA)}{cosA}}\\=\frac{1+cosA}{1-cosA}\\=RHS$

$\frac{tan^{2}A}{(secA-1)^{2}}=\frac{1+cosA}{1-cosA}$

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