Question

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

Answer 1

Let usfirstfind the value of left hand side (LHS) that is $\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$ as shown below:

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

Since LHS=RHS,

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