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If \alpha +\beta =60^{\circ}, then the value of cos^{2}\alpha +cos^{2}\beta -cos\alpha\ cos\beta  is

Option: 1

\frac{3}{4}


Option: 2

-\frac{3}{4}


Option: 3

\frac{4}{3}


Option: 4

-\frac{4}{3}


Answers (1)

best_answer

Answer: Option (a)  \frac{3}{4}

Explanation: 

Given that,

cos^{2}\alpha +cos^{2}\beta -cos\alpha\ cos\beta

\alpha +\beta =60^{\circ}

Multiplying by cos on both sides,

cos\left (\alpha +\beta \right ) =cos\ 60^{\circ}

Expanding the above term to get,

cos\ \alpha \ cos\ \beta -sin\ \alpha \ sin\ \beta =\frac{1}{2}

cos\ \alpha \ cos\ \beta - \frac{1}{2}=sin\ \alpha \ sin\ \beta

Squaring on both sides to get,

\left (cos\ \alpha \ cos\ \beta - \frac{1}{2} \right )^{2}=\left (sin\ \alpha \ sin\ \beta \right )^{2}

  cos^{2}\alpha \ cos^{2}\beta+\frac{1}{4}-2\times cos\ \alpha \ cos\beta \times \frac{1}{2} =sin^{2}\alpha \ sin^{2}\beta

cos^{2}\alpha \ cos^{2}\beta+\frac{1}{4}-cos\ \alpha \ cos\beta =\left ( 1-cos^{2}\alpha \right )\left ( 1-cos^{2}\beta \right )

cos^{2}\alpha \ cos^{2}\beta+\frac{1}{4}-cos\ \alpha \ cos\beta =1-cos^{2}\beta -cos^{2}\alpha +cos^{2}\alpha \ cos^{2}\beta

Canceling and rearranging the terms to get,

cos^{2}\alpha +cos^{2}\beta -cos\alpha\ cos\beta=1-\frac{1}{4}

cos^{2}\alpha +cos^{2}\beta -cos\alpha\ cos\beta=\frac{3}{4}

 

 

Posted by

SANGALDEEP SINGH

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