# Problem-1

prove that ${\frac {1-\tan ^{2}A}{1+\tan ^{2}A}}=1-2\sin ^{2}A$ ## Interpretation of problems

1. It is to prove the problem based on trigonometric identities
2. the function of one trigonometric ratio is relates to other

## Concept development

Develop the skill of proving problem based trigonometric identity

Problem solving

## Pre Knowledge require

1. Idea about trignometric ratios
2. Idea about trignometric identities

## Methos Of Solutions

### Generalisation By Verification

When A=60° LHS=$\displaystyle \frac{1-\tan^2 60°}{1+\tan^2 60°}$
=${\frac {1-{({\sqrt {3}})}^{2}}{1+{({\sqrt {3}})}^{2}}}$ =${\frac {1-3}{1+3}}$ =${\frac {-2}{4}}$ =${\frac {-1}{2}}$ -----(1)
RHS=$1-2\sin ^{2}A$ =$\displaystyle 1-2\sin^260°$
$1-2{({\frac {\sqrt {3}}{2}})}^{2}$ =$1-2({\frac {3}{4}})$ =${\frac {4-2(3)}{4}}$ =${\frac {-1}{2}}$ ------(2)
from eqn1 & eqn2
$\displaystyle \frac{1-\tan^2 60°}{1+\tan^2 60°}$ =$\displaystyle 1-2\sin^260°$
By Generalisation
${\frac {1-\tan ^{2}A}{1+\tan ^{2}A}}=1-2\sin ^{2}A$ ### By Deductive Proof

LHS=${\frac {1-\tan ^{2}A}{1+\tan ^{2}A}}$ =${\frac {1-{\frac {\sin ^{2}A}{\cos ^{2}A}}}{\sec ^{2}A}}$ =${\frac {[\cos ^{2}A-\sin ^{2}A]}{cos^{2}A}}\cos ^{2}A$ =$1-\sin ^{2}A-\sin ^{2}A$ =$1-2\sin ^{2}A$ =RHS