Question

If 3 tan A = 4, then value of $\sqrt{\frac{1-sinA}{1+cosA}}$ is

Answer 1

To find:- $\sqrt{\frac{1-\sin A}{1+\cos A}}$

Given:- $3\tan A=4$

$\Rightarrow \tan A=\frac{4}{3}.....\left(1\right)$

As we know that,

$\tan A=\frac{\text{Perpendicular}}{\text{Base}}.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$ we get

Perpendicular $\left(p\right)=4$

Base $\left(b\right)=3$

Let $h$ be the hypotenuse,

Now, by pythagoras theorem,

${\left(\text{hypotenuse}\right)}^2={\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2$

$\Rightarrow h^2={\left(4\right)}^2+{\left(3\right)}^2$

$\Rightarrow h^2=16+9$

$\Rightarrow h=\sqrt{25}=5$

Therefore,

$\sin \theta =\frac{p}{h}=\frac{4}{5}$

$\Rightarrow 1-\sin A=1-\frac{4}{5}=\frac{5-4}{5}=\frac{1}{5}$

$\cos \theta =\frac{b}{h}=\frac{3}{5}$

$\Rightarrow 1+\cos A=1+\frac{3}{5}=\frac{5+3}{5}=\frac{8}{5}$

Therefore,

$\sqrt{\frac{1-\sin A}{1+\cos A}}$

$=\sqrt{\frac{\left(\frac{1}{5}\right)}{\left(\frac{8}{5}\right)}}$

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

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

Thus the value of $\sqrt{\frac{1-\sin A}{1+\cos A}}$ is $\frac{1}{2\sqrt{2}}$.

Hence the correct answer is $\frac{1}{2\sqrt{2}}$.