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Question: How do you verify \( \dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\sin x+\cos x \) ?...

How do you verify cos⁡x1−tan⁡x+sin⁡x1−cot⁡x=sin⁡x+cos⁡x\dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\sin x+\cos x ?

Explanation

Solution

Hint : We have sum of two terms in the left-hand side of cos⁡x1−tan⁡x+sin⁡x1−cot⁡x=sin⁡x+cos⁡x\dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\sin x+\cos x . We multiply cos⁡x\cos x with cos⁡x1−tan⁡x\dfrac{\cos x}{1-\tan x} and sin⁡x\sin x with sin⁡x1−cot⁡x\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 a2−b2=(a+b)(a−b){{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\cos x to the numerator and denominator of cos⁡x1−tan⁡x\dfrac{\cos x}{1-\tan x} .
We know tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} . This gives tan⁡x.cos⁡x=sin⁡x\tan x.\cos x=\sin x .
So,
cos⁡x1−tan⁡x×cos⁡xcos⁡x=cos⁡2xcos⁡x−sin⁡x\dfrac{\cos x}{1-\tan x}\times \dfrac{\cos x}{\cos x}=\dfrac{{{\cos }^{2}}x}{\cos x-\sin x} .
We multiply sin⁡x\sin x to the numerator and denominator of sin⁡x1−cot⁡x\dfrac{\sin x}{1-\cot x} .
We know
cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x} . This gives cot⁡x.sin⁡x=cos⁡x\cot x.\sin x=\cos x .
So,
sin⁡x1−cot⁡x×sin⁡xsin⁡x=sin⁡2xsin⁡x−cos⁡x\dfrac{\sin x}{1-\cot x}\times \dfrac{\sin x}{\sin x}=\dfrac{{{\sin }^{2}}x}{\sin x-\cos x} .
The equation becomes
cos⁡x1−tan⁡x+sin⁡x1−cot⁡x=cos⁡2xcos⁡x−sin⁡x+sin⁡2xsin⁡x−cos⁡x\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.
cos⁡x1−tan⁡x+sin⁡x1−cot⁡x=cos⁡2xcos⁡x−sin⁡x−sin⁡2xcos⁡x−sin⁡x\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
cos⁡x1−tan⁡x+sin⁡x1−cot⁡x=cos⁡2x−sin⁡2xcos⁡x−sin⁡x\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 a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) . We now apply the identity theorem for the term cos⁡2x−sin⁡2x{{\cos }^{2}}x-{{\sin }^{2}}x . We assume the values a=cos⁡x,b=sin⁡xa=\cos x,b=\sin x .
Applying the theorem, we get
cos⁡2x−sin⁡2x=(cos⁡x+sin⁡x)(cos⁡x−sin⁡x){{\cos }^{2}}x-{{\sin }^{2}}x=\left( \cos x+\sin x \right)\left( \cos x-\sin x \right) .
The equation becomes
cos⁡x1−tan⁡x+sin⁡x1−cot⁡x=(cos⁡x+sin⁡x)(cos⁡x−sin⁡x)(cos⁡x−sin⁡x)\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 (cos⁡x−sin⁡x)\left( \cos x-\sin x \right) from both denominator and numerator.
The equation becomes
cos⁡x1−tan⁡x+sin⁡x1−cot⁡x=(cos⁡x+sin⁡x)\dfrac{\cos x}{1-\tan x}+\dfrac{\sin x}{1-\cot x}=\left( \cos x+\sin x \right) .
Thus verified cos⁡x1−tan⁡x+sin⁡x1−cot⁡x=(cos⁡x+sin⁡x)\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 (cos⁡x−sin⁡x)\left( \cos x-\sin x \right) from both denominator and numerator is (cos⁡x−sin⁡x)≠0\left( \cos x-\sin x \right)\ne 0 . No domain is given for the variable xx . The value of tan⁡x≠0\tan x\ne 0 is essential. The simplified condition will be x≠nπ,n∈Zx\ne n\pi ,n\in \mathbb{Z} .
We also have the multiple angle theorem of cos⁡2x−sin⁡2x=cos⁡2x{{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x .