Question

prove that
$tan {}^{2}a - \tan {}^{2}b = { \sin }^{2}a - { \sin }^{2}b \div { \cos}^{2}a \: { \cos}^{2}b$
​

Answer 1

Step-by-step explanation:

solving left hand side

We have

( tan a - Tan b ) ( tan a + tan b )

(sin a / cos a - sin b / cos b ) ( sin a / cos a + sin b / cos b )

so

( sin² a - sin² b ) / cos² a cos ² b

thus

LHS = RHS