Question

1-tan2A/1+tan2A=1-2sin2A

Answer 1

To prove:

$\bf{\dfrac{1 - tan^{2}A}{1 + tan^{2}A} = 1 - 2sin^{2}A}$

Identities used:

• sec A = 1 ÷ cos A
• tan A = sin A ÷ cos A
• sin² A + cos² A = 1

Solution :-

Taking LHS,

$\sf{\dfrac{1 - tan^{2}A}{1 + tan^{2}A} }$

$= \: \sf{\dfrac{1 - \dfrac{sin^{2}A}{cos^{2}A}}{1 + \dfrac{sin^{2}A}{cos^{2}A}}}$

$\sf{On \: taking \: LCM,}$

$= \: \sf{\dfrac{\dfrac{cos^{2}A - sin^{2}A}{cos^{2}A}}{\dfrac{sin^{2}A + cos^{2}A}{cos^{2}A}}}$

$= \: \sf{\dfrac{cos^{2}A - sin^{2}A}{cos^{2} A} \times \dfrac{cos^{2} A}{1}}$

$= \: \sf{cos^{2}A - sin^{2}A}$

$= \: \sf{(1 - sin^{2}A) - sin^{2}A}$

$= \sf{1 - sin^{2} A - sin^{2} A}$

$= \boxed{\sf{1 - 2 sin^{2}A}}$

Hence Proved