Question

Prove that$(secA+\tan A)(1-\sin A)=\cos A$

Answer 1

As we have $(secA+\tan A)(1-\sin A)=\cos A$

Taking LHS $(secA+\tan A)(1-\sin A)$

As we know that $secA=\frac{1}{\cos A}and\tan A=\frac{\sin A}{\cos A}$

So, $\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\left(1-\sin A\right)$

$$\begin{gathered} \Rightarrow \left(\frac{(1+\sin A)(1-\sin A)}{\cos A}\right) \\ \Rightarrow \left(\frac{1-{\sin }^{2}A}{\cos A}\right) \\ \Rightarrow \frac{{\cos }^{2}A}{\cos A}[1-{\sin }^{2}A={\cos }^{2}A] \\ \Rightarrow \cos A \end{gathered}$$

Therefore, LHS $=$RHS

Hence Proved.