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