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If a \cos 2 \theta+b \sin 2 \theta=c has \alpha and \beta as its roots, then prove that \tan \alpha+\tan \beta=2 b /(a+c) [Hint: Use the identities \cos 2 \theta=\left(\left(1-\tan ^{2} \theta\right) /\left(1+\tan ^{2} \theta\right)\right. and \left.\sin 2 \theta=2 \tan \theta /\left(1+\tan ^{2} \theta\right)\right]
CBSE | Class 11 | Maths | NCERT Exemplar | Trigonometric Functions
If a \cos 2 \theta+b \sin 2 \theta=c has \alpha and \beta as its roots, then prove that \tan \alpha+\tan \beta=2 b /(a+c) [Hint: Use the identities \cos 2 \theta=\left(\left(1-\tan ^{2} \theta\right) /\left(1+\tan ^{2} \theta\right)\right. and \left.\sin 2 \theta=2 \tan \theta /\left(1+\tan ^{2} \theta\right)\right]

Solution:

As per the question,

a \cos 2 \theta+b \sin 2 \theta=c

\alpha and \beta are the roots of the equation.

Now using the multiple angles formula,

It is known that,

\begin{array}{l} \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \text { and } \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta} \\ \therefore a\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)-c=0 \\ \Rightarrow a\left(1-\tan ^{2} \theta\right)+2 b \tan \theta-c\left(1+\tan ^{2} \theta\right)=0 \\ \Rightarrow(-c-a) \tan ^{2} \theta+2 b \tan \theta-c+a=0 \ldots(i) \end{array}

It is known that,

Sum of all the roots of a quadratic equation, a x^{2}+b x+c=0 is given by

(-b / a)

So,

\tan \alpha+\tan \beta=-2 b /-(c+a)=2 b /(c+a)

As a result, \tan \alpha+\tan \beta=2 b /(c+a)

i

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