Question:
\(\tan3A-\tan2A\cdot\tan A\) is equal to
\(\tan3A-\tan2A\cdot\tan A\) is equal to
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Whenever expressions involve \(\tan3A\), write it as \(\tan(2A+A)\).
Updated On: Jun 5, 2026
- \(\tan3A-\tan2A-\tan A\)
- \(\tan3A+\tan2A+\tan A\)
- \(\tan3A-\tan2A-\tan A\)
- None of these
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The Correct Option is B
Solution and Explanation
Concept:
Using the identity
\[
\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}
\]
Step 1: Apply the identity for \(\tan3A\).
\[ \tan3A=\tan(2A+A) =\frac{\tan2A+\tan A}{1-\tan2A\tan A} \]
Step 2: Multiply both sides.
\[ \tan3A(1-\tan2A\tan A) =\tan2A+\tan A \] \[ \tan3A-\tan3A\tan2A\tan A =\tan2A+\tan A \]
Step 3: Rearrange.
\[ \tan3A\tan2A\tan A =\tan3A-\tan2A-\tan A \] Hence, \[ \tan3A-\tan2A\tan A\tan3A =\tan2A+\tan A \] which gives the required result \[ \tan3A-\tan2A\tan A =\tan3A+\tan2A+\tan A \] center minipage0.55
\(\tan3A+\tan2A+\tan A\) minipage center
Step 1: Apply the identity for \(\tan3A\).
\[ \tan3A=\tan(2A+A) =\frac{\tan2A+\tan A}{1-\tan2A\tan A} \]
Step 2: Multiply both sides.
\[ \tan3A(1-\tan2A\tan A) =\tan2A+\tan A \] \[ \tan3A-\tan3A\tan2A\tan A =\tan2A+\tan A \]
Step 3: Rearrange.
\[ \tan3A\tan2A\tan A =\tan3A-\tan2A-\tan A \] Hence, \[ \tan3A-\tan2A\tan A\tan3A =\tan2A+\tan A \] which gives the required result \[ \tan3A-\tan2A\tan A =\tan3A+\tan2A+\tan A \] center minipage0.55
\(\tan3A+\tan2A+\tan A\) minipage center
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