# Question d3713

Jun 7, 2017

see explanation

#### Explanation:

Q : $\frac{\sin x + 1}{\cos} x = \left(\frac{\left(\tan x + \sec x\right) - 1}{\left(\tan x + 1\right) - \sec x}\right) \cdot \left(\frac{\left(\tan x + \sec x\right) + 1}{\left(\tan x + 1\right) + \sec x}\right)$

Let we take RHS to prove LHS,

$\left(\frac{\left(\tan x + \sec x\right) - 1}{\left(\tan x + 1\right) - \sec x}\right) \cdot \left(\frac{\left(\tan x + \sec x\right) + 1}{\left(\tan x + 1\right) + \sec x}\right) = \frac{{\left(\tan x + \sec x\right)}^{2} - 1}{{\left(\tan x + 1\right)}^{2} - {\sec}^{2} x}$

=(tan^2 x + 2tan x sec x + sec ^2 x -1) / (tan^2 x + 2tan x + 1 - sec^2 x 

rearrange,
=(tan^2 x + 2tan x sec x +( sec ^2 x -1)) / (tan^2 x + 1 + 2tan x - (sec^2 x) 

replace ${\sec}^{2} x - 1 = {\tan}^{2} x , {\sec}^{2} x = {\tan}^{2} x + 1$ in the equation
=(tan^2 x + 2tan x sec x + tan^2 x) / (tan^2 x + 1 + 2tan x - (tan^2 x + 1) #

$= \frac{2 {\tan}^{2} x + 2 \tan x \sec x}{2 \tan x}$

$= \tan x + \sec x = \sin \frac{x}{\cos} x + \frac{1}{\cos} x$

$= \frac{\sin x + 1}{\cos} x$ $\to$ proved