NIMCET 2013 Question Paper with solutions

Question : 40 of 120

Marks: +1, -0

The value of tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ is:

Solution:

∵\tan A-\cot A={\sin^2A-\cos^2A}/{\cos A\sin A}

={-2\cos2A}/{\sin 2A}=-2\cot2A

Adding and subtracting cot θ in the equation, we get

\tanθ+2\tan2θ+4\tan4θ+8\cot8θ

=\tanθ-\cotθ+2\tan2θ+4\tan4θ+8\cot8θ+\cotθ

=-2\cot2θ+2\tan2θ+4\tan4θ+8\cot8θ+\cotθ

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