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Question: How do you prove \( \dfrac{{\cos x}}{{1 - \sin x}} = \sec x + \tan x \) ?...

How do you prove cos⁡x1−sin⁡x=sec⁡x+tan⁡x\dfrac{{\cos x}}{{1 - \sin x}} = \sec x + \tan x ?

Explanation

Solution

Hint : In order to prove the given we will consider LHS and divide each terms by cos⁡x\cos x , and we will apply basic trigonometric identities like 1cos⁡x=sec⁡x\dfrac{1}{{\cos x}} = \sec x , sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x , sec⁡2x−tan⁡2x=1{\sec ^2}x - {\tan ^2}x = 1 and algebraic identity a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) . Then, evaluate and prove the given

** Complete step-by-step answer** :
We need to prove cos⁡x1−sin⁡x=sec⁡x+tan⁡x\dfrac{{\cos x}}{{1 - \sin x}} = \sec x + \tan x
Now, let us consider the LHS,
LHS= cos⁡x1−sin⁡x\dfrac{{\cos x}}{{1 - \sin x}}
Let us divide the terms by cos⁡x\cos x ,
=cos⁡xcos⁡x1cos⁡x−sin⁡xcos⁡x= \dfrac{{\dfrac{{\cos x}}{{\cos x}}}}{{\dfrac{1}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}}}
We know that 1cos⁡x=sec⁡x\dfrac{1}{{\cos x}} = \sec x and sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x , then we have,
=1sec⁡x−tan⁡x= \dfrac{1}{{\sec x - \tan x}}
Now, we also know that sec⁡2x−tan⁡2x=1{\sec ^2}x - {\tan ^2}x = 1
Therefore, we have,
=sec⁡2x−tan⁡2xsec⁡x−tan⁡x= \dfrac{{{{\sec }^2}x - {{\tan }^2}x}}{{\sec x - \tan x}}
We know that a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
Thus, LHS =(sec⁡x+tan⁡x)(sec⁡x−tan⁡x)sec⁡x−tan⁡x= \dfrac{{\left( {\sec x + \tan x} \right)\left( {\sec x - \tan x} \right)}}{{\sec x - \tan x}}
Now, by cancelling the common terms, we have,
LHS =sec⁡x+tan⁡x= \sec x + \tan x =RHS
Hence proved that cos⁡x1−sin⁡x=sec⁡x+tan⁡x\dfrac{{\cos x}}{{1 - \sin x}} = \sec x + \tan x .

Alternate method:
Now, let us consider the LHS,
LHS= cos⁡x1−sin⁡x\dfrac{{\cos x}}{{1 - \sin x}}
Let us multiply the numerator and the denominator by 1+sin⁡x1 + \sin x ,
=cos⁡x1−sin⁡x×1+sin⁡x1+sin⁡x= \dfrac{{\cos x}}{{1 - \sin x}} \times \dfrac{{1 + \sin x}}{{1 + \sin x}}
=cos⁡x(1+sin⁡x)(1−sin⁡x)(1+sin⁡x)= \dfrac{{\cos x\left( {1 + \sin x} \right)}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}
By a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) , we have,
=cos⁡x(1+sin⁡x)12−sin⁡2x= \dfrac{{\cos x\left( {1 + \sin x} \right)}}{{{1^2} - {{\sin }^2}x}}
Now, by sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 , we have,
=cos⁡x(1+sin⁡x)cos⁡2x= \dfrac{{\cos x\left( {1 + \sin x} \right)}}{{{{\cos }^2}x}}
Now, cancelling cos⁡x\cos x , we have,
=(1+sin⁡x)cos⁡x= \dfrac{{\left( {1 + \sin x} \right)}}{{\cos x}}
=1cos⁡x+sin⁡xcos⁡x= \dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}
We know that 1cos⁡x=sec⁡x\dfrac{1}{{\cos x}} = \sec x and sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x , then we have,
LHS =sec⁡x+tan⁡x= \sec x + \tan x =RHS
Hence proved that cos⁡x1−sin⁡x=sec⁡x+tan⁡x\dfrac{{\cos x}}{{1 - \sin x}} = \sec x + \tan x .
So, the correct answer is “ cos⁡x1−sin⁡x=sec⁡x+tan⁡x\dfrac{{\cos x}}{{1 - \sin x}} = \sec x + \tan x ”.

Note : In this question it is important to note that whenever we come across these kinds of questions, we always start from the more complex side. This is because it is a lot easier to eliminate terms to make a complex function simple than to find ways to introduce terms to make a simple function complex. Take one step, watch one step. Manipulation of identities and formulas as per need is the most important aspect here.