Question

Probe that. tan^2 theta/(sec theta+1)=sec theta-1​

Answer 1

Answer:

Step-by-step explanation:

Tan²θ/Secθ + 1

=> Sec²θ - 1 / Secθ + 1     (∵ Sec²θ - Tan²θ = 1)

=> (Secθ + 1)(Secθ - 1) / Secθ + 1

=> Secθ - 1

=> R.H.S

Hence Proved.

Answer 2

Question :

$\mathsf{prove \: that } \: \frac {\tan^2 \theta}{ \sec\theta + 1 } = \sec \theta - 1$

Solution :

$\implies\frac{ \tan^2\theta }{\sec\theta + 1}$

$\implies \frac{ \sec^2 \theta - 1}{ \sec\theta + 1 } \: \: \: \: \: \: \: ( \because\sec^2 \theta - \tan^2\theta = 1)$

$\implies \: \frac{( \sec\theta + 1)( \sec\theta - 1) }{( \sec\theta + 1)} \: \: \: \: \: ( \because \: ({a}^{2} - {b}^{2} ) = (a + b)(a - b))$

$\implies (\sec\theta - 1)\implies RHS$

$</p><p>\mathtt{Hence \: proved}$