Question

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

Answer 1

Hint: We have sum of two terms in the left-hand side of $ \dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\sin x+\cos x $ . We multiply $ \cos x $ with $ \dfrac{\cos x}{1-\tan x} $ and $ \sin x $ with $ \dfrac{\sin x}{1-\cot x} $ on both numerator and denominator. Then we add them as the denominators are the same. Then we use the identity of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ to factor the numerator. We also need to mention the conditions.

Complete step-by-step answer:
We multiply $ \cos x $ to the numerator and denominator of $ \dfrac{\cos x}{1-\tan x} $ .
We know $ \tan x=\dfrac{\sin x}{\cos x} $ . This gives $ \tan x.\cos x=\sin x $ .
So,
$ \dfrac{\cos x}{1-\tan x}\times \dfrac{\cos x}{\cos x}=\dfrac{{{\cos }^{2}}x}{\cos x-\sin x} $ .
We multiply $ \sin x $ to the numerator and denominator of $ \dfrac{\sin x}{1-\cot x} $ .
We know
$ \cot x=\dfrac{\cos x}{\sin x} $ . This gives $ \cot x.\sin x=\cos x $ .
So,
$ \dfrac{\sin x}{1-\cot x}\times \dfrac{\sin x}{\sin x}=\dfrac{{{\sin }^{2}}x}{\sin x-\cos x} $ .
The equation becomes
$ \dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\dfrac{{{\cos }^{2}}x}{\cos x-\sin x}+\dfrac{{{\sin }^{2}}x}{\sin x-\cos x} $ .
To make the denominators the same we take a negative sign common.
 $ \dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\dfrac{{{\cos }^{2}}x}{\cos x-\sin x}-\dfrac{{{\sin }^{2}}x}{\cos x-\sin x} $ .
The addition gives
$ \dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{\cos x-\sin x} $
The numerator is in the form of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ . We now apply the identity theorem for the term $ {{\cos }^{2}}x-{{\sin }^{2}}x $ . We assume the values $ a=\cos x,b=\sin x $ .
Applying the theorem, we get
$ {{\cos }^{2}}x-{{\sin }^{2}}x=\left( \cos x+\sin x \right)\left( \cos x-\sin x \right) $ .
The equation becomes
$ \dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\dfrac{\left( \cos x+\sin x \right)\left( \cos x-\sin x \right)}{\left( \cos x-\sin x \right)} $ .
We can now eliminate the $ \left( \cos x-\sin x \right) $ from both denominator and numerator.
The equation becomes
$ \dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\left( \cos x+\sin x \right) $ .
Thus verified $ \dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\left( \cos x+\sin x \right) $ .

Note: It is important to remember that the condition to eliminate the $ \left( \cos x-\sin x \right) $ from both denominator and numerator is $ \left( \cos x-\sin x \right)\ne 0 $ . No domain is given for the variable $ x $ . The value of $ \tan x\ne 0 $ is essential. The simplified condition will be $ x\ne n\pi ,n\in \mathbb{Z} $ .
We also have the multiple angle theorem of $ {{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x $ .