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Find the value: $\tan 3 A - \tan 2 A - \tan A$

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\tan Atan 2 Atan 3 A

\tan 3 Atan 2 Atan A

3 \tan 3 Atan 4 Atan A

2 \tan 3 Atan 2 Atan A

answer is A.

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Detailed Solution

Given the trigonometric expression is,

\tan 3 A - \tan 2 A - \tan A

It is known that,

\tan x = \frac{\sin x}{\cos x}

…………. (1)
So,

\tan (x + y) = \frac{\sin (x + y)}{\cos (x + y)}

…………. (2)
Now it is known that,

\sin (x + y) = \sin xcos y + \sin ycos x

………… (3)

\cos (x + y) = \cos xcos y - \sin xsin y

………… (4)
Put these values in equation 2,

\tan (x + y) = \frac{\sin xcos y + \sin ycos x}{\cos xcos y - \sin xsin y}

Divide numerator and denominator with term,

cosx cosy,

\tan (x + y) = \frac{\frac{si n xco s y}{cosx cosy} + \frac{si n yco s x}{cosx cosy}}{\frac{co s xco s y}{cosx cosy} - \frac{si n xsi n y}{cosx cosy}}

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

…………… (5)
Thus, from the above expression we can write,

\Rightarrow \tan (A + 2 A) = \frac{\tan A + \tan 2 A}{1 - \tan Atan 2 A}

(here, x=A, y=2A)

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

\Rightarrow \tan 3 A - \tan Atan 2 Atan 3 A = \tan A + \tan 2 A

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

This is the required answer.
Option 1 is correct.

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