Question

Prove that (secA+tanA)(1-sinA)=cosA

Answer 1

LHS:
((1/cosA)+(sinA/cosA))*(1-sinA)
=((1+sinA)(1-sinA))/cosa
=1-sin^2(A)/cosA
=cos^2(A)/cosA
=cosA
=RHS

Answer 2

Answer and Explanation:

To prove : $(\sec A+\tan A)(1-\sin A)=\cos A$

Proof :

Taking LHS,

$LHS=(\sec A+\tan A)(1-\sin A)$

Write, $\sec A=\frac{1}{\cos A}\ , \tan A=\frac{\sin A}{\cos A}$

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

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

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

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

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

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

$LHS=\cos A$

$LHS=RHS$

Hence proved.