Question

Prove that
1+ sin 2 A- cos 2 A/1+ sin 2 A+ cos2 A=tan A​

Answer 1

We have to prove,

$\dfrac{1+\sin(2A)-\cos(2A)}{1+\sin(2A)+\cos(2A)}=\tan A$

So,

$\begin{aligned}&\text{LHS}\\\\\implies\ \ &\dfrac{1+\sin(2A)-\cos(2A)}{1+\sin(2A)+\cos(2A)}\\\\\implies\ \ &\dfrac{(1+\sin(2A)-\cos(2A))\sin(2A)}{(1+\sin(2A)+\cos(2A))\sin(2A)}\\\\\implies\ \ &\dfrac{(1+\sin(2A)-\cos(2A))\sin(2A)}{(1+\cos(2A))\sin(2A)+\sin^2(2A)}\\\\\implies\ \ &\dfrac{(1+\sin(2A)-\cos(2A))\sin(2A)}{(1+\cos(2A))\sin(2A)+1-\cos^2(2A)}\quad[\because\ \sin^2x=1-\cos^2x]\end{aligned}$

$\begin{aligned}\\\\\implies\ \ &\dfrac{(1+\sin(2A)-\cos(2A))\sin(2A)}{(1+\cos(2A))\sin(2A)+(1+\cos(2A))(1-\cos(2A))}\\\\\implies\ \ &\dfrac{(1+\sin(2A)-\cos(2A))\sin(2A)}{(1+\cos(2A))(1+\sin(2A)-\cos(2A))}\\\\\implies\ \ &\dfrac{\sin(2A)}{1+\cos(2A)}\\\\\implies\ \ &\dfrac{2\sin A\cos A}{2\cos^2A}\quad[\because\ \sin(2A)=2\sin A\cos A\quad;\quad 1+\cos(2A)=2\cos^2A]\\\\\implies\ \ &\tan A\\\\\implies\ \ &\text{RHS}\end{aligned}$

Hence Proved!