Question

If tana and tanß be the roots of x ^ 2 - px + q = 0 then find cos 2 * (alpha + beta)​

Answer 1

Step-by-step explanation:

We have,

$\tt{ \blue{tan( \alpha )} \: \: \: and \: \: \: \blue{ tan( \beta ) } \: \: \: are \: \: \: the \: \: \: roots \: \: \: of \: \: \: \green{ {x}^{2} - px + q = 0 }}$

So,

$\sf{ \bullet \: \purple{ Sum \: \: of \: \: roots, \: }tan( \alpha ) + tan( \beta ) = p }$

$\sf{ \bullet \: \purple{ Product \: \: of \: \: roots, \: }tan( \alpha ) \cdot tan( \beta ) = q }$

Now,

$\sf{tan( \alpha + \beta ) = \dfrac{tan( \alpha ) + tan( \beta ) }{1 - tan( \alpha ) \cdot tan( \beta ) }}$

$\sf{ \implies \: tan( \alpha + \beta ) = \dfrac{p }{1 - q }}$

So,

$\sf{cos \{2( \alpha + \beta ) \}}$

$\sf{ = \dfrac{1 - tan^{2} ( \alpha + \beta ) }{1 + tan^{2} ( \alpha + \beta ) } }$

$\sf{ = \dfrac{1 - \bigg( \dfrac{p}{1 - q} \bigg) ^{2} }{1 + \bigg( \dfrac{p}{1 - q} \bigg) ^{2} } }$

$\sf{ = \dfrac{ \dfrac{(1 - q) ^{2} - p^{2} }{(1 - q)^{2} } }{ \dfrac{(1 - q) ^{2} + p^{2} }{(1 - q)^{2} }} }$

$\sf{ = \dfrac{ (1 - q) ^{2} - p^{2} }{(1 - q) ^{2} + p^{2} } }$

$\sf{ = \dfrac{ 1 + q ^{2} - 2q - p^{2} }{1 + q ^{2} - 2q + p^{2} } }$