If 3 tan A = 4 then prove that

(i)

$\mathrm{LHS}=\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}$

$=\sqrt{\frac{\left(\frac{1}{\cos \theta}-\frac{1}{\sin \theta}\right)}{\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)}}$

$=\sqrt{\frac{\left(\frac{\sin \theta-\cos \theta}{\sin \theta \cos \theta}\right)}{\left(\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\right)}}$

$=\sqrt{\frac{\left(\frac{\sin \theta-\cos \theta}{\sin \theta}\right)}{\left(\frac{\sin \theta+\cos \theta}{\sin \theta}\right)}}$

$=\sqrt{\frac{\left(\frac{\sin \theta}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)}{\left(\frac{\sin \theta}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right)}}$

$=\sqrt{\frac{1-\cot \theta}{1+\cot \theta}}$

$=\sqrt{\frac{\left(1-\frac{3}{4}\right)}{\left(1+\frac{3}{4}\right)}}$

$=\sqrt{\frac{\left(\frac{1}{4}\right)}{\left(\frac{7}{4}\right)}}$

$=\sqrt{\frac{1}{7}}$

$=\frac{1}{\sqrt{7}}$

$=$ RHS

(ii)

Given : $3 \tan A=4$

$\Rightarrow \tan A=\frac{4}{3}$

Since, $\tan A=\frac{P}{B}$

$\Rightarrow P=4$ and $B=3$

Using Pythagoras theorem,

$P^{2}+B^{2}=H^{2}$

$\Rightarrow 4^{2}+3^{2}=H^{2}$

$\Rightarrow H^{2}=16+9$

$\Rightarrow H^{2}=25$

$\Rightarrow H=5$

Therefore,

$\sin A=\frac{P}{H}=\frac{4}{5}$

$\cos A=\frac{B}{H}=\frac{3}{5}$

Now,

$\sqrt{\frac{1-\sin A}{1+\cos A}}=\sqrt{\frac{1-\frac{4}{5}}{1+\frac{3}{5}}}$

$=\sqrt{\frac{\frac{5-4}{\frac{5}{\frac{5+3}{5}}}}{5}}$

$=\sqrt{\frac{\frac{1}{5}}{\frac{8}{5}}}$

$=\sqrt{\frac{1}{8}}$

$=\frac{1}{2 \sqrt{2}}$

Hence, $\sqrt{\frac{1-\sin A}{1+\cos A}}=\frac{1}{2 \sqrt{2}}$.