If a 2 +b 2 =c has and as its roots, then prove that + = 2b a+b . [Hint: Use the identities 2 =1- ^2 1+ ^2 and 2 =2 1+ ^2 ].

Given that: a 2 +b 2 =c...(i) [1- ^2 1+ ^2 ]+b [2 1+ ^2 ]=c [ 2 =1- ^2 1+ ^2 , 2 =2 1+ ^2 ] -a ^2 +2b =c(1+ ^2 ) -a ^2 +2b =c + c ^2 -a ^2 +2b -c ^2 -c=0 -(a+c) ^2 +2b +(a-c)=0 (a + c) ^2 -2b +(c-a)=0...(ii) Since and are the roots of equation (i) we have and are the roots of (ii) + = -(-2b) a + c [sum of roots of a quadratic equation ax^2+bx+c=0 is -b a ] + = 2b a + c . Hence proved.

If a 2 +b 2 =c has and as its roots, then prove that + = 2b a+b .…

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Question

If $\text{a}\cos2\theta+\text{b}\sin2\theta=\text{c}$ has $\alpha$ and $\beta$ as its roots, then prove that $\tan\alpha+\tan\beta=\frac{2\text{b}}{\text{a+b}}.$
[Hint: Use the identities $\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big].$

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Answer

Given that: $\text{a}\cos2\theta+\text{b}\sin2\theta=\text{c}...(\text{i})$
$\Rightarrow\text{a}\Big[\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big]+\text{b}\Big[\frac{2\tan\theta}{1+\tan^2\theta}\Big]=\text{c}$ $\Big[\because\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta},\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big]$
$\Rightarrow\text{a}-\text{a}\tan^2\theta+2\text{b}\tan\theta=\text{c}(1+\tan^2\theta)$
$\Rightarrow\text{a}-\text{a}\tan^2\theta+2\text{b}\tan\theta=\text{c + c}\tan^2\theta$
$\Rightarrow\text{a}-\text{a}\tan^2\theta+2\text{b}\tan\theta-\text{c}\tan^2\theta-\text{c}=0$
$\Rightarrow-(\text{a}+\text{c})\tan^2\theta+2\text{b}\tan\theta+(\text{a}-\text{c})=0$
$\Rightarrow(\text{a + c})\tan^2\theta-2\text{b}\tan\theta+(\text{c}-\text{a})=0...(\text{ii})$
Since $\alpha$ and $\beta$ are the roots of equation (i) we have $\tan\alpha$ and $\tan\beta$ are the roots of (ii)
$\Rightarrow\tan\alpha+\tan\beta=\frac{-(-2\text{b})}{\text{a + c}}$ [sum of roots of a quadratic equation $\text{ax}^2+\text{bx}+\text{c}=0$ is $\frac{-\text{b}}{\text{a}}$]
$\Rightarrow\tan\alpha+\tan\beta=\frac{2\text{b}}{\text{a + c}}.$ Hence proved.

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