Question

Answer this............

Answer 1

Solution : (Instead of θ, I use A)

_____________________________________________________________

Given :

To Prove :

$\frac{Tan A}{Sec A - 1} = \frac{Tan A + Sec A + 1}{Tan A + Sec A - 1}$

_____________________________________________________________

Proof :

Assuming that they are equal,.

⇒ $\frac{Tan A}{Sec A - 1} = \frac{Tan A + Sec A + 1}{Tan A + Sec A - 1}$

⇒ $(Tan A)(Tan A + Sec A - 1) = (Tan A + Sec A + 1)(Sec A - 1)$

⇒ $(Tan A)(Tan A) +(Tan A)(Sec A) + (Tan A)(-1) = (Tan A)(Sec A - 1) + (Sec A)(Sec A - 1) + (Sec A - 1)(1)$

⇒ $Tan^2 A + Tan A Sec A - Tan A = Tan A Sec A - Tan A + Sec^2 A - Sec A + Sec A - 1$

⇒ $Tan^2 A = Sec^2 A - 1$

We know that,

Sec² A - Tan² A = 1

Hence,

Sec² A = 1 + Tan² A

______________


⇒ Tan² A = (Tan² A + 1) - 1

⇒ Tan² A = Tan² A

⇒ LHS + RHS,.

⇒ Hence, proved,.

_____________________________________________________________

                                              Hope it Helps !!

⇒ Mark as Brainliest,.

Answer 2

$here \: is \: your \: answer \\ hope \: this \: helps \: you$