Question

Prove that \[\dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}}\].

Answer 1

Hint: We prove the equality of LHS to RHS by solving each side separately. Rationalize the fraction having secant and tangent in the denominator. Use the property of trigonometry i.e. \[1 + {\tan ^2}x = {\sec ^2}x\] to cancel out possible terms. Use the values of trigonometric functions in terms of sine and cosine to reach the final answer.
* Rationalizing a fraction is done by multiplying the fraction by a fraction that has both numerator and denominator as the term conjugate to the denominator of the given fraction.

Complete step by step answer:
We have to prove \[\dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}}\]
Solve LHS and RHS of the equation separately.
LHS:
Since, LHS of the equation is \[\dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}}\]
Rationalize the first fraction by multiplying both numerator and denominator by \[\sec x + \tan x\]
\[ \Rightarrow \dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \dfrac{1}{{\sec x - \tan x}} \times \dfrac{{\sec x + \tan x}}{{\sec x + \tan x}} - \dfrac{1}{{\cos x}}\]
\[ \Rightarrow \dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \dfrac{{\sec x + \tan x}}{{(\sec x - \tan x)(\sec x + \tan x)}} - \dfrac{1}{{\cos x}}\]
Use the property \[(a - b)(a + b) = {a^2} - {b^2}\] to write the denominator of the first fraction
\[ \Rightarrow \dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \dfrac{{\sec x + \tan x}}{{{{\sec }^2}x - {{\tan }^2}x}} - \dfrac{1}{{\cos x}}\]
Since, we know\[1 + {\tan ^2}x = {\sec ^2}x\].
Substitute the value of \[{\sec ^2}x\]in the denominator.
\[ \Rightarrow \dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \dfrac{{\sec x + \tan x}}{{1 + {{\tan }^2}x - {{\tan }^2}x}} - \dfrac{1}{{\cos x}}\]
Cancel the same terms in the denominator having opposite signs i.e. \[{\tan ^2}x\] and \[ - {\tan ^2}x\]
\[ \Rightarrow \dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \dfrac{{\sec x + \tan x}}{1} - \dfrac{1}{{\cos x}}\]
Substitute the value of second fraction using \[\dfrac{1}{{\cos x}} = \sec x\]
\[ \Rightarrow \dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \sec x + \tan x - \sec x\]
Cancel the same terms in the denominator having opposite signs i.e. \[\sec x\] and \[\sec x\].
\[ \Rightarrow \dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\cos x}} = \tan x\] ………...… (1)
RHS:
Since, RHS of the equation is \[\dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}}\]
Rationalize the second fraction by multiplying both numerator and denominator by \[\sec x - \tan x\]
\[ \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} = \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} \times \dfrac{{\sec x - \tan x}}{{\sec x - \tan x}}\]
\[ \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} = \dfrac{1}{{\cos x}} - \dfrac{{\sec x - \tan x}}{{(\sec x + \tan x)(\sec x - \tan x)}}\]
Use the property \[(a - b)(a + b) = {a^2} - {b^2}\] to write the denominator of the first fraction
\[ \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} = \dfrac{1}{{\cos x}} - \dfrac{{\sec x - \tan x}}{{{{\sec }^2}x - {{\tan }^2}x}}\]
Since, we know\[1 + {\tan ^2}x = {\sec ^2}x\].
Substitute the value of \[{\sec ^2}x\]in the denominator.
\[ \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} = \dfrac{1}{{\cos x}} - \dfrac{{\sec x - \tan x}}{{1 + {{\tan }^2}x - {{\tan }^2}x}}\]
Cancel the same terms in the denominator having opposite signs i.e. \[{\tan ^2}x\] and \[ - {\tan ^2}x\]
\[ \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} = \dfrac{1}{{\cos x}} - \dfrac{{\sec x - \tan x}}{1}\]
Substitute the value of second fraction using \[\dfrac{1}{{\cos x}} = \sec x\]
\[ \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} = \sec x - (\sec x - \tan x)\]
\[ \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} = \sec x - \sec x + \tan x\]
Cancel the same terms in the denominator having opposite signs i.e. \[\sec x\] and \[\sec x\].
\[ \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{1}{{\sec x + \tan x}} = \tan x\] ………….… (2)
From equations (1) and (2),
Values of LHS \[ = \tan x\] and of RHS\[ = \tan x\]
\[\therefore \] LHS \[ = \]RHS

Hence Proved.

Note:
Students might make the mistake of solving the LHS and RHS by taking LCM in the starting which will give us a complex solution. Also, many students cross multiply the fractions of both sides to solve the equation, but here we don’t have to solve the equation, we have to prove left hand side to right hand side.