Question

If sec(theta+alpha)+sec(theta-alpha)=2sec theta and cos alpha is not equal to 1 then show that cos theta=+- root2 cos alpha/2

Answer 1

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Answer 2

It is showed  that  $cos\theta = \± \sqrt{2 \cos\frac{\alpha }{2} }$.

Step-by-step explanation:

Given:

If $sec(\theta+\alpha ) +sec(\theta -\alpha ) =2 sec\theta$.

$cos\alpha \neq 1$.

To Find:

It is to be showed that $cos\theta = \± \sqrt{2 \cos\frac{\alpha }{2} }$.

Identity Used:

$Cos A + Cos B = 2 Cos \frac{(A+B)}{2} . Cos\frac{(A-B)}{2}$

$2Cos A \ Cos B =Cos (A + B) - Cos (A -B)$

$cos2\theta = 2 cos^{ 2} \theta-1$

Solution:

As given- $sec(\theta+\alpha ) +sec(\theta -\alpha ) =2 sec\theta$.

$sec(\theta+\alpha ) +sec(\theta -\alpha ) =2 sec\theta$

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

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

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

$\frac{ 2\times 2 cos\theta cos\alpha }{cos2 \theta+ cos2\alpha } =\frac{2}{cos \theta}$

$2cos^{2} \theta\ cos\alpha = cos2\theta + cos2\alpha$

$2cos^{2} \theta\ cos\alpha = 2cos^{2}\theta-1 + 2cos^{2}\alpha -1$

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

$cos^{2} \theta =\frac{cos^{2}\alpha -1}{cos\alpha-1}$

$cos^{2} \theta =cos\alpha +1$

$cos^{2} \theta =2cos^{2} {(\frac{\alpha }{2} )}$

$cos\theta = \± \sqrt{2 \cos\frac{\alpha }{2} }$

Hence, it is It is showed  that  $cos\theta = \± \sqrt{2 \cos\frac{\alpha }{2} }$.