The value of tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα is
cot2nα
2ntan2nα
0
cotα
The explanation for the correct option:
Compute the required value:
The given trigonometric expression: tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα.
Now, cotA-tanA=cotA-1cotA∵tanθ=1cotθ
⇒cotA-tanA=cot2A-1cotA⇒cotA-tanA=2×cot2A-12cotA⇒cotA-tanA=2cot2A∵cot2θ=cot2θ-12cotθ
Thus, tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα=tanα-cotα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα+cotα
⇒tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα=-2cot2α+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα+cotα⇒tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα=-4cot4α+4tan4α+...+2n-1tan2n-1α+2ncot2nα+cotα⇒tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα=-8cot8α+...+2n-1tan2n-1α+2ncot2nα+cotα⇒tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα=-2n-1cot2n-1α+2n-1tan2n-1α+2ncot2nα+cotα⇒tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα=-2ncot2nα+2ncot2nα+cotα⇒tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα=cotα
Hence, the value of tanα+2tan2α+4tan4α+...+2n-1tan2n-1α+2ncot2nα is cotα.
Hence, option D is the correct answer.