If cos α +cos β +cos γ =0=sin α +sin β +sin γ then cos2 α +cos2 β +cos2 γ equals Options: A . &nbsp 2cos( α + β + γ ) B . &nbsp cos2( α + β + γ ) C . &nbsp 0 D . &nbsp 1

If cosα+cosβ+cosγ=0=sinα+sinβ+sinγ then c

Question -

If cos $α$ +cos $β$ +cos $γ$ =0=sin $α$ +sin $β$ +sin $γ$ then

cos2 $α$ +cos2 $β$ +cos2 $γ$ equals

Options:

A . 2cos(

α

β

γ

)

B . cos2(

α

β

γ

)

C . 0

D . 1

Answer: Option C
:
C

cos $α$ +cos $β$ +cos $γ$ =0 ......(i)

sin $α$ +sin $β$ +sin $γ$ =0 .......(ii)

Let a=cos $α$ +isin $α$ , b=cos $β$ +isin $β$

c=cos $γ$ +isin $γ$

$\Rightarrow$ a+b+c=0 .......(iii)

$\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$

= ${[c o s α + i s i n α]}^{- 1}$ + ${[c o s β + i s i n β]}^{- 1}$ + ${[c o s γ + i s i n γ]}^{- 1}$

=cos $α$ -isin $α$ +cos $β$ -isin $β$ +cos $γ$ -isin $γ$

$\Rightarrow$ ab+bc+ca=0 .....(iv) [by (i) and (ii)]

Squaring both sides of equations (iii),

We get $a^{2}$ + $b^{2}$ + $c^{2}$ +2ab+2bc+2ca=0

or $a^{2}$ + $b^{2}$ + $c^{2}$ [by (iv)]

${[c o s α + i s i n α]}^{2}$ + ${[c o s β + i s i n β]}^{2}$ + ${[c o s γ + i s i n γ]}^{2}$ =0

$\Rightarrow$ (cos2 $α$ +cos2 $β$ +cos2 $γ$ ) + i(sin2 $α$ +sin2 $β$ +sin2 $γ$ )

Equating the real part to zero, we have

$\Rightarrow c o s 2 α + c o s 2 β + c o s 2 γ = 0$

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