Tan -2tan 2-4tan 4-8cot 8-IIT 1988- MP PET 1991-

=tan α+2tan 2α+4tan 4α+8cot 8α tan α+2tan 2α+4[sin 4α/cos 4α+2cos 8α/sin 8α] =tan α+2tan 2α+4[cos 4α cos 8α+sin 4α sin 8α+cos 4α cos 8α/sin 8α cos 4α] =tan α+ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

tan α+2tan 2α...

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]

$t a n α$

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$t a n 2 α$

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$c o t α$

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$c o t 2 α$

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Solution

The correct option is C
$c o t α$

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

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

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

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

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

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

Suggest Corrections

tan α+2tan 2α+4tan 4α+8cot 8α= [IIT 1988; MP PET 1991]

$t a n α + 2 t a n 2 α + 4 t a n 4 α + 8 c o t 8 α$ =

[IIT 1988; MP PET 1991]