Question

How do you prove the identity $\dfrac{{\tan x + \sec x - 1}}{{\tan x - \sec x + 1}} = \tan x + \sec x$ ?

Answer 1

Hint : In order to prove $\dfrac{{\tan x + \sec x - 1}}{{\tan x - \sec x + 1}} = \tan x + \sec x$ , we will multiply both the numerator and denominator with $\tan x + \sec x + 1$ . Then, simplify it and use trigonometric identity ${\sec ^2}x = {\tan ^2}x + 1$ . And, evaluating it we will get the RHS.

Complete step-by-step answer :
Now, let us consider the LHS of the equation,
$= \dfrac{{\tan x + \sec x - 1}}{{\tan x - \sec x + 1}}$
Let us multiply both the numerator and denominator with $\tan x + \sec x + 1$ ,
$= \dfrac{{\tan x + \sec x - 1}}{{\tan x - \sec x + 1}} \times \dfrac{{\tan x + \sec x + 1}}{{\tan x + \sec x + 1}}$
$= \dfrac{{{{\tan }^2}x + \tan x\sec x + \tan x + \sec x\tan x + {{\sec }^2}x + \sec x - \tan x - \sec x - 1}}{{{{\tan }^2}x + \tan x\sec x + \tan x - \sec x\tan x - {{\sec }^2}x - \sec x + \tan x + \sec x + 1}}$
By cancelling and simplifying, we have,
$= \dfrac{{{{\tan }^2}x + 2\tan x\sec x + {{\sec }^2}x - 1}}{{{{\tan }^2}x - {{\sec }^2}x + 2\tan x + 1}}$
We know from trigonometric identities, ${\sec ^2}x = {\tan ^2}x + 1$ .
By substituting we have,
$= \dfrac{{{{\tan }^2}x + 2\tan x\sec x + {{\tan }^2}x + 1 - 1}}{{{{\tan }^2}x - {{\tan }^2}x - 1 + 2\tan x + 1}}$
$= \dfrac{{2{{\tan }^2}x + 2\tan x\sec x}}{{2\tan x}}$
Take $\tan x$ common out from the numerator.
$= \dfrac{{2\tan x\left( {\tan x + \sec x} \right)}}{{2\tan x}}$
Now, cancel $2\tan x$ as it is common on both numerator and denominator, we have,
$= \tan x + \sec x$
Therefore, LHS=RHS
Hence, $\dfrac{{\tan x + \sec x - 1}}{{\tan x - \sec x + 1}} = \tan x + \sec x$ .
So, the correct answer is “ $\tan x + \sec x$ ”.

Note : Trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles. It is expressed as ratios of sine(sin), cosine(cos), tangent(tan), cotangent(cot), secant(sec), cosecant(cosec) angles