Question

Prove that $\frac{(1-\sin A)}{(1+\sin A)}={(secA-\tan A)}^2$

Answer 1

In $\frac{(1-\sin A)}{(1+\sin A)}={(secA-\tan A)}^2$,

$L.H.S=\frac{\left(1-\sin A\right)}{\left(1+\sin A\right)}$

By rationalising the denominator we get,

$\Rightarrow \frac{\left(1-\sin A\right)}{\left(1+\sin A\right)}\times \frac{\left(1-\sin A\right)}{\left(1-\sin A\right)}$

$\Rightarrow \frac{{\left(1-\sin A\right)}^2}{{\left(1\right)}^2-{\sin }^{2}A}$…………………………..$\left(a+b\right)\left(a-b\right)=a^2-b^2$

$\Rightarrow \frac{{\left(1-\sin A\right)}^2}{1-{\sin }^{2}A}$

$\Rightarrow \frac{{\left(1-\sin A\right)}^2}{{\cos }^{2}A}$………………….$\left(1-{\sin }^{2}A={\cos }^{2}A\right)$

$\Rightarrow {\left(\frac{1-\sin A}{\cos A}\right)}^2$

$\Rightarrow {\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)}^2$

$\Rightarrow {\left(secA-\tan A\right)}^2$…………………….$\left(\frac{1}{\cos \theta }=sec\theta ,\frac{\sin \theta }{\cos \theta }=\tan \theta \right)$

$=R.H.S$.

Hence, it is proved that $\frac{(1-\sin A)}{(1+\sin A)}={(secA-\tan A)}^2$.