Question

If tan.alpha and tan.beta be the roots of x^2 – px + q = 0, then find cos2(alpha + beta).​

Answer 1

Answer:

$\cos ^{2} ( \alpha + \beta ) = \frac{ {(1 - q)}^{2} }{ {p}^{2} + {(1 - q) }^{2} }$

Step-by-step explanation:

tan.alpha and tan.beta

roots of

x^2 – px + q = 0,

$\tan( \alpha ) + \tan( \beta ) = p \\ \tan( \alpha \tan( \beta ) ) = q$

$\tan( \alpha + \beta ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha) \tan( \beta ) }$

$\tan( \alpha + \beta ) = \frac{p}{1 - q}$

$\tan ^{2} ( \alpha + \beta ) = \frac{ {p}^{2} }{ {(1 - q)}^{2} }$

$\sec^{2} ( \alpha + \beta ) = 1 + \tan ^{2} ( \alpha + \beta )$

$\sec ^{2} ( \alpha + \beta ) = 1 + \frac{ {p}^{2} }{ {(1 - q)}^{2} }$

$\sec^{2} ( \alpha + \beta ) = \frac{ {(1 - q)}^{2} + {p}^{2} }{ {(1 - q)}^{2} }$

$\cos ^{2} ( \alpha + \beta ) = \frac{1}{ \sec ^{2} ( \alpha + \beta ) }$

$\cos ^{2} ( \alpha + \beta ) = \frac{ {(1 - q)}^{2} }{ {p}^{2} + {(1 - q) }^{2} }$

Answer 2

answer is given in the attachment given below very clearly