Question

Prove that:-
(1+tan^3/1+tan) + tan -sec^2 =0
​

Answer 1

To prove

$\sf\frac{1 + tan^3\theta}{1 + tan\theta} + tan\theta - sec^2\theta = 0$

Solution

Using a³ + b³ = (a + b)(a² + b² - ab), we can write

1 + tan³∅ = (1 + tan∅)(1 + tan²∅ - tan∅)

Hence, the expression becomes

$\sf\frac{(1 + tan\theta)(1 + tan^2\theta - tan\theta)}{(1 + tan\theta)} + tan\theta - sec^2\theta$

Cancelling (1 + tan∅) from both numerator and denominator,

$\sf 1 + tan^2\theta - tan\theta + tan\theta - sec^2\theta$

Cancel out tan∅ from negative tan∅

Also, we know that 1 + tan²∅ = sec²∅

So,

$\sf sec^2\theta - sec^2\theta$

= 0

Hence Proved.