Question

sinA.tanA/1-cosA=1+secA

Answer 1

sina.tana/1-cosa
=(sina.sina/cosa)/1-cosa
=(sin^2a/cosa)/1-cosa
=(1-cos^2a/cosa)/1-cosa
={(1-cosa)(1+cosa)/cosa}/1-cosa
=1+cosa/cosa
=seca+1
it is the prove
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Answer 2

Answer and explanation:

To prove : $\frac{\sin A\cdot\tan A}{1-\cos A}=1+\sec A$

Proof :

Taking LHS,

$LHS=\frac{\sin A\cdot\tan A}{1-\cos A}$

$LHS=\frac{\sin A\cdot\frac{\sin A}{\cos A}}{1-\cos A}$

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

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

$LHS=\frac{\frac{(1-\cos A)(1+\cos A)}{\cos A}}{1-\cos A}$

$LHS=\frac{1+\cos A}{\cos A}$

$LHS=\sec A+1$

$LHS=RHS$

Hence proved.