Question

Show that ${\tan }^{2}A-{\sin }^{2}A={\tan }^{2}A{\sin }^{2}A$

Answer 1

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

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

Since LHS=RHS,

Hence, ${\tan }^{2}A-{\sin }^{2}A={\tan }^{2}A{\sin }^{2}A$.