t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α = [IIT 1988; MP PET 1991] Options: A . &nbsp t a n α B . &nbsp t a n 2 α C . &nbsp c o t α D . &nbsp c o t 2 α

Answer: Option C : C = t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α t a n α + 2 t a n 2 α + 4 [ s i n 4 α c o s 4 α + 2 c o s 8 α s i n 8 α ] = t a n α + 2 t a n 2 α + 4 [ c o s 4 α c o s 8 α + s i n 4 α s i n 8 α + c o s 4 α c o s 8 α s i n 8 α c o s 4 α ] = t a n α + 2 t a n 2 α + 4 [ c o s 4 α + c o s 4 α c o s 8 α s i n 8 α c o s 4 α ] = t a n α + 2 t a n 2 α + 4 [ c o s 4 α ( 1 + c o s 8 α ) c o s 4 α s i n 8 α ] = t a n α + 2 t a n 2 α + 4 [ 2 c o s 2 4 α 2 s i n 4 α c o s 4 α ] = t a n α + 2 t a n 2 α + 4 c o t 4 α = t a n α + 2 ( t a n 2 α + 2 c o t 4 α ) = t a n α + 2 [ s i n 2 α c o s 2 α + 2 c o s 4 α s i n 4 α ] = t a n α + 2 [ c o s 2 α ( 1 + c o s 4 α ) s i n 4 α c o s 2 α ] = t a n α + 2 c o t 2 α = s i n α c o s α + 2 c o s 2 α s i n 2 α = c o s α + c o s α c o s 2 α s i n 2 α c o s α = 1 + c o s 2 α s i n 2 α = 2 c o s 2 α 2 s i n α c o s α = c o t α

tan α+2tan 2α+4tan 4α+8cot 8α= [I

Discussion Forum : Trigonometric Functions

Question -

$t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α$ =
[IIT 1988; MP PET 1991]

Options:

A .

t a n α

B .

t a n 2 α

C .

c o t α

D .

c o t 2 α

Answer: Option C
:
C

= $t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α$
$t a n α + 2 t a n 2 α + 4 [\frac{s i n 4 α}{c o s 4 α} + 2 \frac{c o s 8 α}{s i n 8 α}]$
= $t a n α + 2 t a n 2 α + 4 [\frac{c o s 4 α c o s 8 α + s i n 4 α s i n 8 α + c o s 4 α c o s 8 α}{s i n 8 α c o s 4 α}]$
= $t a n α + 2 t a n 2 α + 4 [\frac{c o s 4 α + c o s 4 α c o s 8 α}{s i n 8 α c o s 4 α}]$
= $t a n α + 2 t a n 2 α + 4 [\frac{c o s 4 α (1 + c o s 8 α)}{c o s 4 α s i n 8 α}]$
= $t a n α + 2 t a n 2 α + 4 [\frac{2 c o s^{2} 4 α}{2 s i n 4 α c o s 4 α}]$
= $t a n α + 2 t a n 2 α + 4 c o t 4 α$
= $t a n α + 2 (t a n 2 α + 2 c o t 4 α)$
= $t a n α + 2 [\frac{s i n 2 α}{c o s 2 α} + 2 \frac{c o s 4 α}{s i n 4 α}]$
= $t a n α + 2 [\frac{c o s 2 α (1 + c o s 4 α)}{s i n 4 α c o s 2 α}]$
= $t a n α + 2 c o t 2 α = \frac{s i n α}{c o s α} + \frac{2 c o s 2 α}{s i n 2 α}$
= $\frac{c o s α + c o s α c o s 2 α}{s i n 2 α c o s α}$
= $\frac{1 + c o s 2 α}{s i n 2 α} = \frac{2 c o s^{2} α}{2 s i n α c o s α} = c o t α$

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