Question

Find the value of
tan3A. tan2A. tanA

• A. tan2A-tanA+tan3A
• B. tan2A+tanA+tan3A
• C. tan3A-tan2A-tanA
• D. tan3A+tan2A-tanA

Answer 1

The correct option is C tan3A-tan2A-tanA

We know,

$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

$\therefore \tan 3A=\tan (2A+A)=\frac{\tan 2A+\tan A}{1-\tan 2A\tan A}$

$\Rightarrow \tan 3A(1-\tan 2A\tan A)=tan2A+\tan A$

$\Rightarrow \tan 3A-\tan 2A-\tan A=\tan 3A\tan 2A\tan A$

Hence, proved.