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If a Cos 2 X + B Sin 2 X = C Has α and β as Its Roots, Then Prove that (Ii) Tan α Tan β = C − a C + a - Mathematics

Numerical

If \[a \cos2x + b \sin2x = c\]  has α and β as its roots, then prove that

(ii)  \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]

 

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Solution

Given: \[a \cos2x + b \sin2x = c\]

\[\Rightarrow a\left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) + b\left( \frac{2\text{ tan } x}{1 + \tan^2 x} \right) - c = 0\]
\[ \Rightarrow a\left( 1 - \tan^2 x \right) + 2b \text{ tan } x - c\left( 1 + \tan^2 x \right) = 0\]
\[ \Rightarrow a - a \tan^2 x + 2b \text{ tan } x - c - c \tan^2 x = 0\]
\[ \Rightarrow \left( a + c \right) \tan^2 x - 2b \text{ tan } x + \left( c - a \right) = 0 . . . . . \left( 1 \right)\]

This a quadratic equation in terms of \[\tan^2 x\] . 

It is given that α and β are the roots of the given equation, so tan α and tan β are the roots of (1).
Since tan α and tan β are the roots of the equation

\[\left( a + c \right) \tan^2 x - 2b \text{ tan } x + \left( c - a \right) = 0\]. Therefore,
(ii) \[\tan\alpha \tan\beta = \frac{c - a}{a + c} \left( \text{ Product of roots }  = \frac{c}{a} \right)\]

\[\text{ Or } \tan\alpha \tan\beta = \frac{c - a}{c + a}\] . 

 
 

 

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 44.2 | Page 30
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