Question

If $ y = \dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}} $ , prove that $ \dfrac{{dy}}{{dx}} + {y^2} + 1 = 0 $

Answer 1

Hint: We need to convert the equation in a form having an only single term, in this it will be tan x. Then using the quotient rule of derivation we need to find the value of $ \dfrac{{dy}}{{dx}} $ and $ {y^2} $ then accordingly $ \dfrac{{dy}}{{dx}} + {y^2} + 1 = 0 $ will be proved.

Complete step-by-step answer:
The given equation is having both sin x and cos x, we can proceed directly by using the quotient rule of derivation but it will lead to confusion so it’s better to convert the equation into a form having either sin x or cos x or any other trigonometric form. Hence, we need to divide the numerator and the divider both by cos x.
So, $ y = \dfrac{{\dfrac{{\left( {\cos x - \sin x} \right)}}{{\cos x}}}}{{\dfrac{{\left( {\cos x + \sin x} \right)}}{{\cos x}}}} $
 $ y = \dfrac{{\left( {\dfrac{{\cos x}}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{{\cos x}}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}} \right)}} $
And since $ \dfrac{{\sin x}}{{\cos x}} = \tan x $ ;
 $ \Rightarrow y = \dfrac{{\left( {1 - \tan x} \right)}}{{\left( {1 + \tan x} \right)}} $
Now, we need to find $ \dfrac{{dy}}{{dx}} $ using the Quotient rule of derivatives. But at first we need to know and memorize the Quotient rule of derivatives which is as follow:
 $ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} $
Hence, applying the above rule we will be getting as below:
\[\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{\left( {1 - \tan x} \right)}}{{\left( {1 + \tan x} \right)}}} \right)\]
 $ \dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 + \tan x} \right)\dfrac{d}{{dx}}(1 - \tan x) - \left( {1 - \tan x} \right)\dfrac{d}{{dx}}(1 + \tan x)}}{{{{\left( {1 + \tan x} \right)}^2}}} $
Since, $ \dfrac{d}{{dx}}(1 - \tan x) = - {\sec ^2}x $ and $ \dfrac{d}{{dx}}(1 + \tan x) = {\sec ^2}x $
So,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 + \tan x} \right)\left( { - {{\sec }^2}x} \right) - \left( {1 - \tan x} \right)\left( {{{\sec }^2}x} \right)}}{{{{\left( {1 + \tan x} \right)}^2}}} $
\[\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - {{\sec }^2}x - \tan x.{{\sec }^2}x - {{\sec }^2}x + \tan x.{{\sec }^2}x}}{{{{\left( {1 + \tan x} \right)}^2}}}\]
\[\dfrac{{dy}}{{dx}} = \dfrac{{ - 2{{\sec }^2}x}}{{{{\left( {1 + \tan x} \right)}^2}}}\]
Since, $ {\sec ^2}x = (1 + {\tan ^2}x) $
\[\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2\left( {1 + {{\tan }^2}x} \right)}}{{{{\left( {1 + \tan x} \right)}^2}}}\]
\[\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2\left( {1 + {{\tan }^2}x} \right)}}{{{{\left( {1 + \tan x} \right)}^2}}} = \dfrac{{ - 2 - 2{{\tan }^2}x}}{{{{\left( {1 + \tan x} \right)}^2}}}\]
Now, we need to find the value of $ {y^2} $ where $ y = \dfrac{{\left( {1 - \tan x} \right)}}{{\left( {1 + \tan x} \right)}} $
So,
 $ \Rightarrow {y^2} = \dfrac{{{{\left( {1 - \tan x} \right)}^2}}}{{{{\left( {1 + \tan x} \right)}^2}}} = \dfrac{{1 + {{\tan }^2}x - 2\tan x}}{{{{\left( {1 + \tan x} \right)}^2}}} $
Now, we have to prove that $ \dfrac{{dy}}{{dx}} + {y^2} + 1 = 0 $ , so putting the derived values of $ \dfrac{{dy}}{{dx}} $ and $ {y^2} $ in the given equation we will be having as below
 \[\Rightarrow \dfrac{{ - 2 - 2{{\tan }^2}x}}{{{{\left( {1 + \tan x} \right)}^2}}} + \dfrac{{1 + {{\tan }^2}x - 2\tan x}}{{{{\left( {1 + \tan x} \right)}^2}}} + 1 = \dfrac{{ - 2 - 2{{\tan }^2}x + 1 + {{\tan }^2}x - 2\tan x + {{\left( {1 + \tan x} \right)}^2}}}{{{{\left( {1 + \tan x} \right)}^2}}}\]
 $ = \dfrac{{ - 2 - 2{{\tan }^2}x + 1 + {{\tan }^2}x - 2\tan x + 1 + {{\tan }^2}x + 2\tan x}}{{{{\left( {1 + \tan x} \right)}^2}}} $
 $ = \dfrac{{ - 2 - 2{{\tan }^2}x + 2 + 2{{\tan }^2}x}}{{{{\left( {1 + \tan x} \right)}^2}}} $
Hence proved, $ \dfrac{{dy}}{{dx}} + {y^2} + 1 = 0 $

Note: Kindly don’t confuse or forget while using the Quotient rule of derivatives because if we don’t remember it we won’t be able to proceed further and will be in complete mess. Additionally, we also need to remember the derivatives of all the basic trigonometric forms.