Question

If cos (x+a) cos (x-a)=1/2, then prove that
tan^2 x=(1-tan^2 a)/(1+3tan^2 a)​

Answer 1

Step-by-step explanation:

We have,

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

$\implies \{ \cos(x) \cos( \alpha ) - \sin(x) \sin( \alpha ) \} \{ \cos(x) \cos( \alpha ) + \sin(x) \sin( \alpha ) \}= \frac{1}{2}$

$\implies \cos^{2} (x) \cos ^{2} ( \alpha ) - \sin ^{2} (x) \sin ^{2} ( \alpha ) = \frac{1}{2}$

$\implies \cos^{2} (x) \cos ^{2} ( \alpha ) - \{1 - \cos ^{2} (x) \} \{1 - \cos^{2} ( \alpha ) \} = \frac{1}{2}$

$\implies \cos^{2} (x) \cos ^{2} ( \alpha ) - \{1 - \cos ^{2} (x) - \cos^{2} ( \alpha ) + \cos ^{2} (x) \cos^{2} ( \alpha ) \} = \frac{1}{2}$

$\implies \cos^{2} (x) \cos ^{2} ( \alpha ) - 1 + \cos ^{2} (x) + \cos^{2} ( \alpha ) - \cos ^{2} (x) \cos^{2} ( \alpha ) = \frac{1}{2}$

$\implies - 1 + \cos ^{2} (x) + \cos^{2} ( \alpha ) = \frac{1}{2}$

$\implies \cos ^{2} (x) = \frac{1}{2} + 1 - \cos^{2} ( \alpha )$

$\implies \cos ^{2} (x) = \frac{3}{2} - \cos^{2} ( \alpha )$

$\implies \cos ^{2} (x) = \frac{3- 2\cos^{2} ( \alpha )}{2}$

$\implies \sec^{2} (x) = \frac{2}{3- 2\cos^{2} ( \alpha )}$

$\implies \sec^{2} (x) - 1 = \frac{2}{3- 2\cos^{2} ( \alpha )} - 1$

$\implies \tan^{2} (x) = \frac{2 - 3 + 2 \cos ^{2} ( \alpha ) }{3- 2\cos^{2} ( \alpha )}$

$\implies \tan^{2} (x) = \frac{ - 1 + 2 \cos ^{2} ( \alpha ) }{3- 2\cos^{2} ( \alpha )}$

$\implies \tan^{2} (x) = \frac{ - ( \sin^{2} ( \alpha ) + \cos ^{2} ( \alpha )) + 2 \cos ^{2} ( \alpha ) }{3 ( \sin^{2} ( \alpha ) + \cos ^{2} ( \alpha )) - 2\cos^{2} ( \alpha )}$

$\implies \tan^{2} (x) = \frac{ - \sin^{2} ( \alpha ) - \cos ^{2} ( \alpha ) + 2 \cos ^{2} ( \alpha ) }{3 \sin^{2} ( \alpha ) + 3\cos ^{2} ( \alpha ) - 2\cos^{2} ( \alpha )}$

$\implies \tan^{2} (x) = \frac{ - \sin^{2} ( \alpha ) + \cos ^{2} ( \alpha ) }{3 \sin^{2} ( \alpha ) + \cos ^{2} ( \alpha ) }$

$\implies \tan^{2} (x) = \frac{ \frac{- \sin^{2} ( \alpha ) + \cos ^{2} ( \alpha ) }{ \cos^{2} ( \alpha ) }}{ \frac{3 \sin^{2} ( \alpha ) + \cos ^{2} ( \alpha ) }{ \cos^{2} ( \alpha ) }}$

$\implies \tan^{2} (x) = \frac{ - \tan^{2} ( \alpha ) + 1}{3 \tan^{2} ( \alpha ) + 1 }$

$\implies \tan^{2} (x) = \frac{1 - \tan^{2} ( \alpha ) }{1 + 3 \tan^{2} ( \alpha ) }$