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

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

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