Question

How do you simplify $ \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} $ ?

Answer 1

Hint: We can solve this type of trigonometric equations by using basic concepts and formulae of trigonometry. Trigonometric identities such as $ {\sin ^2}x + {\cos ^2}x = 1 $ and $ {\sec ^2}x = 1 + {\tan ^2}x $ will be of significant use in solving the problem. We will also use the double angle formula of trigonometric ratios like cosine and sine to get to the final answer.

Complete step-by-step answer:
So, we have, $ \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} $ .
Now, we know the trigonometric identity $ {\sin ^2}x + {\cos ^2}x = 1 $ . Dividing both sides of this equation by $ {\cos ^2}x $ on both sides, we get,
 $ \Rightarrow \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}x}} $
Separating the denominators, we get,
 $ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}x}} $
Now, we know that $ \dfrac{{\sin x}}{{\cos x}} = \tan x $ and $ \dfrac{1}{{\cos x}} = \sec x $ . So, we get,
 $ \Rightarrow {\tan ^2}x + 1 = {\sec ^2}x $
So, we get,
 $ \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \dfrac{{1 - {{\tan }^2}x}}{{{{\sec }^2}x}} $
Again using the trigonometric formulae $ \dfrac{{\sin x}}{{\cos x}} = \tan x $ and $ \dfrac{1}{{\cos x}} = \sec x $ . So, we get,
 $ \Rightarrow \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \dfrac{{1 - \left( {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}{{\left( {\dfrac{1}{{{{\cos }^2}x}}} \right)}} $
Taking LCM in the numerator,
 $ \Rightarrow \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \dfrac{{\left( {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}{{\left( {\dfrac{1}{{{{\cos }^2}x}}} \right)}} $
Cancelling the common factors in numerator and denominator, we get,
 $ \Rightarrow \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = {\cos ^2}x - {\sin ^2}x $
We can leave the final answer at this stage. But, we can also simplify the expression further using the double angle formula of cosine.
We know the double angle formula for cosine is $ \cos 2x = {\cos ^2}x - {\sin ^2}x $ .
So, we get the simplified value of the expression given to us in the question $ \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} $ as $ \cos 2x $ .
So, the correct answer is “ $ \cos 2x $”.

Note: One must know the basic concepts and formulae related to trigonometry such as $ \dfrac{{\sin x}}{{\cos x}} = \tan x $ and $ \dfrac{1}{{\cos x}} = \sec x $ . We should know the double angle formula for cosine in various forms such as $ \cos 2x = \left( {{{\cos }^2}x - {{\sin }^2}x} \right) = \left( {2{{\cos }^2}x - 1} \right) = \left( {1 - 2{{\sin }^2}x} \right) $ so as to simplify any expression related to trigonometry. We must take care of calculations while taking LCM of the denominators so as to be sure of the final simplification of the expression provided to us.