Question

help me with basic trigonometry question

I made some mistake lol yeah...!

please solve it

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Answer 1

We Have :-

sec ( θ + α ) + sec ( θ - α ) = 2 sec θ

To Prove :-

cos θ = √2 cos α / 2

Solution :-

$sec ( \theta + \alpha ) + sec ( \theta - \alpha ) = 2 sec \theta\\\\\frac{1}{cos( \theta + \alpha)} + \frac{1}{cos( \theta - \alpha )} = \frac{2}{cos \theta}$

$\frac{cos ( \theta + \alpha )+cos( \theta - \alpha )}{cos(\theta + \alpha )cos( \theta - \alpha )} = \frac{2}{cos \theta} \\\\\frac{2*2 cos \theta cos \alpha}{cos ( \theta + \alpha + \theta - \alpha ) + cos ( \theta + \alpha - \theta + \alpha )} = \frac{2}{cos \theta} \\\\\frac{2 cos \theta cos \alpha }{cos 2 \theta + cos 2 \alpha} = \frac{1}{cos \theta} \\\\2 cos^{2} \theta cos \alpha = cos 2 \theta + cos 2 \alpha\\\\2 cos^{2} \theta cos \alpha = 2 cos^{2} \theta - 1 + 2 cos^{2} \alpha - 1$

$2 cos^{2} \theta ( cos \alpha - 1)=2 cos^{2} \alpha - 2\\\\cos^{2} \theta = \frac{cos^{2} \alpha - 1}{cos \alpha - 1} = cos \alpha + 1\\ \\cos^{2} \theta = 2 cos ^{2} ( \frac{\alpha}{2} )\\\\cos \theta = \sqrt{2} cos ( \frac{\alpha}{2})$

Hence Proved

Answer 2

Answer:

(*handwriting is worst plz don't mind)

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