# Problem-1

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

## 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

## Methos Of Solutions

### Generalisation By Verification

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

### By Deductive Proof

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