JEE MATHEMATICS MCQ: MULTIPLE AND SUBMULTIPLE ANGLES

1 Hard

1 of 1

Analyzing +4 | -1

For

x \in R

\tan \,\,x + \frac{1}{2}\,\,\tan \,\,\frac{x}{2}\,\, + \,\,\frac{1}{{{2^2}}}\,\,\tan \,\,\frac{x}{{{2^2}}} + \ldots + \frac{1}{{{2^{n - 1}}}}\,\,\tan \,\,\left( {\frac{x}{{{2^{n - 1}}}}} \right)

is equal to

View answer and explanation

Correct option: B

\frac{1}{{{2^{n - 1}}}}\,\,\cot \,\,\left( {\frac{x}{{{2^{n - 1}}}}} \right) - 2\cot \,\,2x

Option B is correct because by using the identity

\tan x = \cot x - 2\cot 2x

repeatedly, we can express the given sum as a telescoping series, which simplifies to the given expression.

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