Question

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

Answer 1

Hint: In order to prove the given we will consider LHS and divide each terms by $ \cos x $ , and we will apply basic trigonometric identities like $ \dfrac{1}{{\cos x}} = \sec x $ , $ \dfrac{{\sin x}}{{\cos x}} = \tan x $ , $ {\sec ^2}x - {\tan ^2}x = 1 $ and algebraic identity $ {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 $ \dfrac{{\cos x}}{{1 - \sin x}} = \sec x + \tan x $
Now, let us consider the LHS,
LHS= $ \dfrac{{\cos x}}{{1 - \sin x}} $
Let us divide the terms by $ \cos x $ ,
  $ = \dfrac{{\dfrac{{\cos x}}{{\cos x}}}}{{\dfrac{1}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}}} $
We know that $ \dfrac{1}{{\cos x}} = \sec x $ and $ \dfrac{{\sin x}}{{\cos x}} = \tan x $ , then we have,
  $ = \dfrac{1}{{\sec x - \tan x}} $
Now, we also know that $ {\sec ^2}x - {\tan ^2}x = 1 $
Therefore, we have,
 $ = \dfrac{{{{\sec }^2}x - {{\tan }^2}x}}{{\sec x - \tan x}} $
We know that $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $
Thus, LHS $ = \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 $ =RHS
Hence proved that $ \dfrac{{\cos x}}{{1 - \sin x}} = \sec x + \tan x $ .

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