Activity-trigonometry problems

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

Skill development

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=Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\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=$1-2\sin ^{2}A$
=Failed to parse (syntax error): {\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
Failed to parse (syntax error): {\displaystyle \frac{1-\tan^2 60°}{1+\tan^2 60°}} =Failed to parse (syntax error): {\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