Question

Prove the following: $\frac{\tan A}{1+secA}-\frac{\tan A}{1-secA}=2\cos ecA$

Answer 1

To prove $\frac{\tan A}{1+secA}-\frac{\tan A}{1-secA}=2\cos ecA$.

consider L.H.S.:

$\begin{array}{rcl}\mathrm{LHS}& =& \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}\\ & =& \frac{\tan A(1-\sec A)-\tan A(1+\sec A)}{(1+secA)(1-secA)}\\ & =& \frac{\tan A-\tan A·\sec A-\tan A-\tan A·\sec A}{1-se{c}^{2}A}\left[\because (a+b)(a-b)=a^2-b^2\right]\\ & =& -\frac{2\tan A·\sec A}{-{\tan }^{2}A}\left[\because {\sec }^{2}A-{\tan }^{2}A=1\right]\\ & =& \frac{2\sec A}{\tan A}\\ & =& \frac{2}{\cos A}\times \frac{\cos A}{\sin A}\left[\because \sec A=\frac{1}{\cos A}\mathrm{and}\tan A=\frac{\sin A}{\cos A}\right]\\ & =& \frac{2}{\sin A}\\ & =& 2\cos ecA\\ & =& \mathrm{RHS}\end{array}$

Thus, $\mathrm{LHS}=\mathrm{RHS}$

Hence, $\frac{\tan A}{1+secA}-\frac{\tan A}{1-secA}=2\cos ecA$ is proved.