Question

Show that $\frac{\tan A-\sin A}{{\sin }^{2}A}=\frac{\tan A}{1+\cos A}$

Answer 1

Let usfirstfind the value of left hand side (LHS) that is $\frac{\tan A-\sin A}{{\sin }^{2}A}$ as shown below:

$\frac{\tan A-\sin A}{{\sin }^{2}A}=\frac{\frac{\sin A}{\cos A}-\sin A}{1-{\cos }^{2}A}(\because \tan x=\frac{\sin x}{\cos x},{\sin }^{2}x=1-{\cos }^{2}x)=\frac{\frac{\sin A-\sin A\cos A}{\cos A}}{(1-\cos A)(1+\cos A)}(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)=\frac{\sin A(1-\cos A)}{\cos A}\times \frac{1}{(1-\cos A)(1+\cos A)}=\tan A\times \frac{1}{(1+\cos A)}(\because \tan x=\frac{\sin x}{\cos x},{\sin }^{2}x=1-{\cos }^{2}x)=\frac{\tan A}{1+\cos A}=RHS$

Since LHS=RHS,

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