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Question

tan α+2tan 2α+4tan 4α+8cot 8α=

[IIT 1988; MP PET 1991]


A

tanα

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B

tan2α

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C

cotα

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D

cot2α

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Solution

The correct option is C

cotα


=tan α+2tan 2α+4tan 4α+8cot 8α
tan α+2tan 2α+4[sin 4αcos 4α+2cos 8αsin 8α]

=tan α+2tan 2α+4[cos 4α cos 8α+sin 4α sin 8α+cos 4α cos 8αsin 8α cos 4α]

=tan α+2tan 2α+4[cos 4α+cos 4α cos 8αsin 8α cos 4α]

=tan α+2tan 2α+4[cos 4α(1+cos 8α)cos 4α sin 8α]

=tan α+2tan 2α+4[2cos2 4α2sin 4α cos 4α]

=tan α+2tan 2α+4cot 4α
=tan α+2(tan 2α+2cot 4α)
=tan α+2[sin 2αcos 2α+2cos 4αsin 4α]
=tan α+2[cos 2α(1+cos 4α)sin 4α cos 2α]
=tan α+2cot 2α=sin αcos α+2cos 2αsin 2α
=cos α+cos α cos 2αsin 2α cos α
=1+cos 2αsin 2α=2cos2α2sin α cos α=cot α


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