Question

Prove the identity

Answer 1

Answer:

Proved

Step-by-step explanation:

The options are not given. So, I will solve from scratch

Given

$\frac{secx-1}{tan x}= \frac{tanx}{secx+1}$

Required

Prove

Multiply the right-hand side by $\frac{secx + 1}{secx + 1}$

$\frac{secx-1}{tan x} * \frac{secx + 1}{secx + 1}= \frac{tanx}{secx+1}$

Apply difference of two squares on the numerator

$\frac{sec^2 x - 1}{(tanx)(secx + 1)} =\frac{tanx}{secx+1}$

In trigonometry:

$tan^2x = sec^2x - 1$

So, we have:

$\frac{tan^2 x}{(tanx)(secx + 1)} =\frac{tanx}{secx+1}$

$\frac{tan x * tan x}{(tanx)(secx + 1)} =\frac{tanx}{secx+1}$

tan x cancels out

$\frac{tan x}{secx + 1} =\frac{tanx}{secx+1}$

Proved