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  • If a cos2θ + b sin 2θ = c has α and β as its roots, then prove that tan α + tan β = 2b/(a + c) [Hint: Use the identities cos 2θ = (( 1 – tan2 θ)/(1 + tan2 θ) and sin 2θ = 2tan θ/(1 + tan2 θ)]

If a \cos2 \theta + b \sin 2 \theta = c  has α and β as its roots, then prove that \tan \alpha + \tan \beta = 2b/(a + c)[Hint: Use the identities \cos 2 \theta = (( 1 - \tan\textsuperscript{2} \theta )/(1 + \tan^2 \theta ) and \sin 2 \theta = 2\tan \theta /(1 + \tan^2 \theta )]

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\\ \text{Given that a}\cos 2 \theta +b\sin 2 \theta =c \ldots \ldots \ldots \left( i \right) ~~~ \\\\ ~ a - a\tan^{2} \theta +2b \tan \theta =c \left( 1+\tan ^{2} \theta \right) ~ \\\\ ~ a - a\tan ^{2} \theta +2b \tan \theta =c+c \tan^{2} \theta \\\\ \left( a+c \right) \tan ^{2} \theta - 2b \tan \theta + \left( c - a \right) =0 \ldots .. \left( ii \right) ~ \\\\

Since \alpha \: \: and \: \: \beta are the roots of the equation i)we have tanα and tanβ are the roots of ii)
   \\ \tan \alpha +\tan \beta = \frac{ - \left( - 2b \right) }{a+c}~~~~ \left[ \text{Sum of roots of a quadratic equation} \right] ~~ \\\\ \tan \alpha +\tan \beta =2b/ \left( a+c \right) \\\\

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