Question

$\left[\frac{1}{(\tan 3A-\tan A)}\right]-\left[\frac{1}{(cot3A-cotA)}\right]=$

• A. $\tan A$
• B. $\tan 2A$
• C. $cotA$
• D. $cot2A$

Answer 1

The correct option is D

$cot2A$

Explanation for the correct answer:

Step 1. Simplify the given expression using trigonometric identities

$\frac{1}{(\tan 3A-\tan A)}-\frac{1}{(cot3A-cotA)}$$=\frac{1}{(\tan 3A-\tan A)}-\frac{1}{\left({\displaystyle \frac{1}{\tan 3A}}-{\displaystyle \frac{1}{\tan A}}\right)}$ $\left[\because tan\theta =\frac{1}{cot\theta }\right]$

$=\frac{1}{(\tan 3A-\tan A)}-\frac{\tan 3A.\tan A}{(\tan A-\tan 3A)}$

$=\frac{1}{(\tan 3A-\tan A)}+\frac{\tan 3A.\tan A}{(\tan 3A-\tan A)}$

$=\frac{\left(1+\tan 3A.\tan A\right)}{(\tan 3A-\tan A)}$

Step 2. Divide numerator and denominator by $1+\tan 3A.\tan A$

$\frac{1}{(\tan 3A-\tan A)}-\frac{1}{(cot3A-cotA)}$$=\frac{1}{\left({\displaystyle \frac{\tan 3A-\tan A}{1+\tan 3A.\tan A}}\right)}$

$=\frac{1}{\tan (3A-A)}$ $\left[\because \frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{A}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{B}}{\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{A}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{B}}\mathbf{=}\mathbf{tan}\mathbf{(}\mathit{A}\mathbf{-}\mathit{B}\mathbf{)}\right]$

$=\frac{1}{\tan 2A}=\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathbf{2}\mathit{A}$

Hence, Option ‘D’ is Correct.