Question

We have to prove that

Prove that $(1+cotA-\cos ecA)(1+\tan A+secA)=2$

Answer 1

Prove the given expression. $(1+cotA-\cos ecA)(1+\tan A+secA)=2$

L.H.S. $=(1+cotA-\cos ecA)(1+\tan A+secA)$

$$\begin{gathered} =\left(1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{1}{\cos A}\right) \\ =\left(\frac{\sin A+\cos A-1}{\sin A}\right)\left(\frac{\cos A+\sin A+1}{\cos A}\right) \\ =\frac{{(\sin A+\cos A)}^2-1^2}{\sin A.\cos A} \\ =\frac{{\sin }^{2}A+{\cos }^{2}A+2\sin A.\cos A-1}{\sin A.\cos A} \\ =\frac{1+2\sin A.\cos A-1}{\sin A.\cos A} \\ =2R.H.S \end{gathered}$$

Hence, L.H.S. = R.H.S. proved.