Question

In any triangle ABC, the simplified form of `(cos2A)/a^2 - (cos2B)/b^2` is ______

विकल्प

• a² - b² 

• `1/(a^2 - b^2)`

• `1/a^2 - 1/b^2`

• a² + b² 

Answer 1

In any triangle ABC, the simplified form of `(cos2A)/a^2 - (cos2B)/b^2` is `underline(1/a^2 - 1/b^2)`.

Explanation:

`(cos2A)/a^2 - (cos2B)/b^2`

= `(1 - 2sin^2A)/a^2 - (1 - 2sin^2B)/b^2`

= `1/a^2 - 1/b^2 - (2sin^2A)/a^2 + (2sin^2B)/b^2`

= `1/a^2 - 1/b^2 - 2((sin^2A)/a^2 - (sin^2B)/b^2)`

= `1/a^2 - 1/b^2` .......`["By sine rule", a/(sinA) = b/(sinB)]`