Question

Tan^2A - tan ^2 B=sin sq A-sin sq B/cos sq A*cos sq B

Answer 1

Correct Question:

Prove that : tan² A - tan² B = ( sin² A - sin² B ) / cos²A. cos² B

Solution:

tan² A - tan² B = ( sin² A - sin² B ) / cos²A. cos² B

Consider LHS

= tan² A - tan² B

Since tan² Φ = sin² Φ / cos² Φ

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

Taking LCM

$\sf = \dfrac{sin^{2} A .cos^{2} B + sin^{2} B.cos ^{2}A}{cos^{2}A cos ^{2} B}$

Using trigonometric identity cos² Φ = 1 - sin² Φ

$\sf = \dfrac{sin^{2} A (1 - sin^{2} B) - sin^{2} B(1 - sin ^{2}A)}{cos^{2}A cos ^{2} B}$

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

Cancelling like terms in the numerator

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

= RHS

Hence proved.