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Question: The expression \[\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)\] is equal to A). \[3\ta...

The expression tan⁡A+tan⁡(60+A)−tan⁡(60−A)\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right) is equal to
A). 3tan⁡3A3\tan 3A
B). tan⁡3A\tan 3A
C). cot⁡3A\cot 3A
D). sin⁡3A\sin 3A

Explanation

Solution

In order to find the solution to the given multiple-choice question that is to find tan⁡A+tan⁡(60+A)−tan⁡(60−A)\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)is equal to which of the given options, simplify the given trigonometric expression with help of following identities of tangent in trigonometry that are: tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}, tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}and tan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x\tan 3x=\dfrac{3\tan x-{{\tan }^{3}}x}{1-3{{\tan }^{2}}x}.

Complete step by step solution:
According to the question, given trigonometric expression in the question is as follows:
tan⁡A+tan⁡(60+A)−tan⁡(60−A)...(1)\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)...\left( 1 \right)
To simplify the above expression, apply one of the trigonometric identities of tangent that istan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y} on tan⁡(60+A)\tan \left( 60+A \right), we get:
⇒tan⁡(60+A)=tan⁡60∘+tan⁡A1−tan⁡60∘tan⁡A\Rightarrow \tan \left( 60+A \right)=\dfrac{\tan {{60}^{\circ }}+\tan A}{1-\tan {{60}^{\circ }}\tan A}
We know that tan⁡60∘=3\tan {{60}^{\circ }}=\sqrt{3}, substituting this value in the above equation, we get:
⇒tan⁡(60+A)=3+tan⁡A1−3tan⁡A...(2)\Rightarrow \tan \left( 60+A \right)=\dfrac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}...\left( 2 \right)
Now apply another trigonometric identity of tangent that istan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y} on tan⁡(60−A)\tan \left( 60-A \right), we get:
⇒tan⁡(60−A)=tan⁡60∘−tan⁡A1+tan⁡60∘tan⁡A\Rightarrow \tan \left( 60-A \right)=\dfrac{\tan {{60}^{\circ }}-\tan A}{1+\tan {{60}^{\circ }}\tan A}
We know that tan⁡60∘=3\tan {{60}^{\circ }}=\sqrt{3}, substituting this value in the above equation, we get:
⇒tan⁡(60+A)=3−tan⁡A1+3tan⁡A...(3)\Rightarrow \tan \left( 60+A \right)=\dfrac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}...\left( 3 \right)
Now with the help of the equation (2)\left( 2 \right) and (3)\left( 3 \right)we will first simplify, tan⁡(60+A)−tan⁡(60−A)\tan \left( 60+A \right)-\tan \left( 60-A \right) as follows:
⇒tan⁡(60+A)−tan⁡(60−A)=3+tan⁡A1−3tan⁡A−3−tan⁡A1+3tan⁡A\Rightarrow \tan \left( 60+A \right)-\tan \left( 60-A \right)=\dfrac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}-\dfrac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}
To simplify it further, solve the terms in the right-hand side of the equation by taking the LCM, we get:
⇒tan⁡(60+A)−tan⁡(60−A)=(3+tan⁡A)(1+3tan⁡A)−(3−tan⁡A)(1−3tan⁡A)(1−3tan⁡A)(1+3tan⁡A)\Rightarrow \tan \left( 60+A \right)-\tan \left( 60-A \right)=\dfrac{\left( \sqrt{3}+\tan A \right)\left( 1+\sqrt{3}\tan A \right)-\left( \sqrt{3}-\tan A \right)\left( 1-\sqrt{3}\tan A \right)}{\left( 1-\sqrt{3}\tan A \right)\left( 1+\sqrt{3}\tan A \right)}
Now the identity (a2−b2)=(a+b)(a−b)\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right) on the denominator of the fraction in the right-hand side of the above equation, we get:
⇒(3+tan⁡A)(1+3tan⁡A)−(3−tan⁡A)(1−3tan⁡A)12−(3tan⁡A)2\Rightarrow \dfrac{\left( \sqrt{3}+\tan A \right)\left( 1+\sqrt{3}\tan A \right)-\left( \sqrt{3}-\tan A \right)\left( 1-\sqrt{3}\tan A \right)}{{{1}^{2}}-{{\left( \sqrt{3}\tan A \right)}^{2}}}
To solve it further, open the brackets with help of multiplication and addition in the above equation, we get:
⇒3+3tan⁡A+tan⁡A+3tan⁡2A−3+3tan⁡A+tan⁡A−3tan⁡2A1−3tan⁡2A\Rightarrow \dfrac{\sqrt{3}+3\tan A+\tan A+\sqrt{3}{{\tan }^{2}}A-\sqrt{3}+3\tan A+\tan A-\sqrt{3}{{\tan }^{2}}A}{1-3{{\tan }^{2}}A}
After solving the terms in the numerator of the above expression we get:
⇒tan⁡(60+A)−tan⁡(60−A)=8tan⁡A1−3tan⁡2A...(4)\Rightarrow \tan \left( 60+A \right)-\tan \left( 60-A \right)=\dfrac{8\tan A}{1-3{{\tan }^{2}}A}...\left( 4 \right)
Now, substituting the value of the equation (4)\left( 4 \right) in the equation (1)\left( 1 \right), we get:
⇒tan⁡A+tan⁡(60+A)−tan⁡(60−A)=tan⁡A+8tan⁡A1−3tan⁡2A\Rightarrow \tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)=\tan A+\dfrac{8\tan A}{1-3{{\tan }^{2}}A}
To simplify it further, solve the terms in the right-hand side of the equation by taking the LCM, we get:
⇒tan⁡A−3tan⁡3A+8tan⁡A1−3tan⁡2A\Rightarrow \dfrac{\tan A-3{{\tan }^{3}}A+8\tan A}{1-3{{\tan }^{2}}A}
After solving the terms in the numerator of the above expression, we get:
⇒3(3tan⁡A−tan⁡3A)1−3tan⁡2A\Rightarrow \dfrac{3\left( 3\tan A-{{\tan }^{3}}A \right)}{1-3{{\tan }^{2}}A}
Now apply one of the trigonometric identities of tangent that is tan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x\tan 3x=\dfrac{3\tan x-{{\tan }^{3}}x}{1-3{{\tan }^{2}}x} in the above expression, we get:
⇒tan⁡A+tan⁡(60+A)−tan⁡(60−A)=3tan⁡3A\Rightarrow \tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)=3\tan 3A
Therefore, tan⁡A+tan⁡(60+A)−tan⁡(60−A)\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right) is equal to 3tan⁡3A3\tan 3Aand hence option (a) is the correct answer.

Note: Students usually get confused and interchange between the sign of the identities tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}and tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}. Key point is to remember the right formula/identity.