Question

Show that:-
1+cosA/1-cosA=tan2A/(secA-1)2​

Answer 1

SOLUTION

$\bf{L\:H\:S}$

$\:\:\:\sf \frac{1+ \cos A}{1-\cos A}$

$=\sf\frac{(1+\cos A)(1-\cos A)}{(1-\cos A)^2}$

$=\sf \frac{(1-\cos^2A)}{{(1-\cos A)^2}}^{}$

$=\sf \frac{sin^2A}{(1-\cos A)^2}\times \frac{\frac{1}{\cos^2A}}{\frac{1}{\cos^2A}}$

$= \sf \frac{tan^2A}{\frac{1}{\cos^2A}-1}$

$\therefore{\underline{\boxed{\sf{\frac{\tan^2A}{(\sec^2A-1)^2}}}}}}=\bf R\:H\:S$

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