Question

CosA+Cos(4π/3-A)+(4π\3+A)=0

Answer 1

Please see the attachment

Answer 2

$\displaystyle \sf{cos \: A + cos \bigg( \frac{4\pi}{3} + A \bigg) +cos \bigg( \frac{4\pi}{3} - A \bigg) = 0} \: \: is \: proved$

Given :

$\displaystyle \sf{cos \: A + cos \bigg( \frac{4\pi}{3} + A \bigg) +cos \bigg( \frac{4\pi}{3} - A \bigg) = 0}$

To find :

To prove the expression

Solution :

Step 1 of 2 :

Write down the given expression to prove

$\displaystyle \sf{cos \: A + cos \bigg( \frac{4\pi}{3} + A \bigg) +cos \bigg( \frac{4\pi}{3} - A \bigg) = 0}$

Step 2 of 2 :

Prove the expression

LHS

$\displaystyle \sf{ = cos \: A + cos \bigg( \frac{4\pi}{3} + A \bigg) +cos \bigg( \frac{4\pi}{3} - A \bigg) }$

$\displaystyle \sf{ = cos \: A + cos ( {240}^{ \circ} + A ) +cos ( {240}^{ \circ} - A ) }$

$\displaystyle \sf{ = cos \: A + 2.\: \: cos \bigg( \frac{{240}^{ \circ} + A + {240}^{ \circ} - A}{2} \bigg) cos \bigg( \frac{{240}^{ \circ} + A - {240}^{ \circ} + A}{2} \bigg) }$

$\displaystyle \sf{ = cos \: A + 2.\: \: cos \bigg( \frac{{240}^{ \circ} + {240}^{ \circ} }{2} \bigg) cos \bigg( \frac{ A + A}{2} \bigg) }$

$\displaystyle \sf{ = cos \: A + 2.\: \: cos \bigg( \frac{{480}^{ \circ} }{2} \bigg) cos \bigg( \frac{ 2A }{2} \bigg) }$

$\displaystyle \sf{ = cos \: A + 2 \: cos {240}^{ \circ} cos A }$

$\displaystyle \sf{ = cos \: A + 2 \: cos (3 \times {90}^{ \circ} -{30}^{ \circ} ) \: cos A }$

$\displaystyle \sf{ = cos \: A - 2 \:sin \: {30}^{ \circ} \: cos A }$

$\displaystyle \sf{ = cos \: A - 2 \times \frac{1}{2} \times cos A }$

$\displaystyle \sf{ = cos \: A - cos A }$

$\displaystyle \sf{ = 0}$

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