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Question: Prove that \[\tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}\]...

Prove that tan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x\tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}

Explanation

Solution

Hint : Here in this question we have to prove the given inequality which is given in this question. This question involves the trigonometric function we should know about the trigonometry ratio. Hence by using the simple calculations we are going to prove the given inequality.

Complete step-by-step answer :
In trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given inequality tan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x\tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}
Now we consider the LHS
⇒tan⁡3x\Rightarrow \tan 3x
The angle of the tangent trigonometry ratio can be written in terms of sum of 2x and x
⇒tan⁡(2x+x)\Rightarrow \tan (2x + x)
By using the sum formula for tangent trigonometry ratio is given by tan⁡(a+b)=tan⁡a+tan⁡b1−tan⁡atan⁡b\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}
So we have
⇒tan⁡2x+tan⁡x1−tan⁡2x.tan⁡x\Rightarrow \dfrac{{\tan 2x + \tan x}}{{1 - \tan 2x.\tan x}}
By the double angle formula tan⁡2x=2tan⁡x1−tan⁡2x\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}, On substituting this term in the above equation
⇒2tan⁡x1−tan⁡2x+tan⁡x1−2tan⁡x1−tan⁡2x.tan⁡x\Rightarrow \dfrac{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} + \tan x}}{{1 - \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}.\tan x}}
On taking LCM for both numerator and denominator terms we have
⇒2tan⁡x+tan⁡x−tan⁡3x1−tan⁡2x1−tan⁡2x−2tan⁡2x1−tan⁡2x\Rightarrow \dfrac{{\dfrac{{2\tan x + \tan x - {{\tan }^3}x}}{{1 - {{\tan }^2}x}}}}{{\dfrac{{1 - {{\tan }^2}x - 2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}}}
On cancelling the terms we get
⇒2tan⁡x+tan⁡x−tan⁡3x1−tan⁡2x−2tan⁡2x\Rightarrow \dfrac{{2\tan x + \tan x - {{\tan }^3}x}}{{1 - {{\tan }^2}x - 2{{\tan }^2}x}}
On simplifying we get
⇒3tan⁡x−tan⁡3x1−3tan⁡2x\Rightarrow \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}
⇒RHS\Rightarrow RHS
Here we have proved LHS = RHS.
So, the correct answer is “Option B”.

Note : The question involves the trigonometric functions and we have to prove the trigonometric function. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities.