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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
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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