Question

prove LHS IS EQUAL TO RHS.

CHALLENGE FOR TOPPERS.​

Answer 1

Step-by-step explanation:

For answer refer to the attachment

Answer 2

We have,

$\tan(45^{\circ}-\theta)=\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}$

So, let $\theta=\dfrac{A}{2}.$ Then we can have,

$\tan\left(45^{\circ}-\dfrac{A}{2}\right)=\dfrac{\cos\left(\dfrac{A}{2}\right)-\sin\left(\dfrac{A}{2}\right)}{\cos\left(\dfrac{A}{2}\right)+\sin\left(\dfrac{A}{2}\right)}$

The RHS can be squared, then be square rooted! The value won't change!

$\tan\left(45^{\circ}-\dfrac{A}{2}\right)=\sqrt{\left(\dfrac{\cos\left(\dfrac{A}{2}\right)-\sin\left(\dfrac{A}{2}\right)}{\cos\left(\dfrac{A}{2}\right)+\sin\left(\dfrac{A}{2}\right)}\right)^2}$

The RHS only is considered below.

$\begin{aligned}&\sqrt{\left(\dfrac{\cos\left(\dfrac{A}{2}\right)-\sin\left(\dfrac{A}{2}\right)}{\cos\left(\dfrac{A}{2}\right)+\sin\left(\dfrac{A}{2}\right)}\right)^2}\\\\\implies\ \ &\sqrt{\dfrac{\left(\cos\left(\dfrac{A}{2}\right)-\sin\left(\dfrac{A}{2}\right)\right)^2}{\left(\cos\left(\dfrac{A}{2}\right)+\sin\left(\dfrac{A}{2}\right)\right)^2}}\end{aligned}$

$\begin{aligned}\implies\ \ &\sqrt{\dfrac{\cos^2\left(\dfrac{A}{2}\right)+\sin^2\left(\dfrac{A}{2}\right)-2\cos\left(\dfrac{A}{2}\right)\sin\left(\dfrac{A}{2}\right)}{\cos^2\left(\dfrac{A}{2}\right)+\sin^2\left(\dfrac{A}{2}\right)+2\cos\left(\dfrac{A}{2}\right)\sin\left(\dfrac{A}{2}\right)}\end{aligned}$

And we know that,

$\bullet\ \sin^2\theta+\cos^2\theta=\cos^2\theta+\sin^2\theta=1\\\\\bullet\ 2\sin\theta\cos\theta=2\cos\theta\sin\theta=\sin(2\theta)$

Hence, again taking $\theta=\dfrac{A}{2},$ we get,

$\begin{aligned}&\sqrt{\dfrac{\cos^2\left(\dfrac{A}{2}\right)+\sin^2\left(\dfrac{A}{2}\right)-2\cos\left(\dfrac{A}{2}\right)\sin\left(\dfrac{A}{2}\right)}{\cos^2\left(\dfrac{A}{2}\right)+\sin^2\left(\dfrac{A}{2}\right)+2\cos\left(\dfrac{A}{2}\right)\sin\left(\dfrac{A}{2}\right)}\\\\\implies\ \ &\sqrt{\dfrac{1-\sin A}{1+\sin A}}\end{aligned}$

Finally,

$\large\boxed{\quad\quad\quad\quad\quad\quad\quad\tan\left(45^{\circ}-\dfrac{A}{2}\right)=\sqrt{\dfrac{1-\sin A}{1+\sin A}}\quad\quad\quad\quad\quad\quad\quad}$

Hence Proved!