Question

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

Answer 1

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

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

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