Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

If 2 Tan α = 3 Tan β , Prove that Tan ( α − β ) = Sin 2 β 5 − Cos 2 β . - Mathematics

Numerical

If $2 \tan \alpha = 3 \tan \beta,$  prove that $\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}$ .

Solution

Given: $2 \tan \alpha = 3 \tan \beta$

$LHS = \frac{tan\alpha - tan\beta}{1 + tan\alpha \times tan\beta}$
$= \frac{\frac{3}{2} \times tan\beta - tan\beta}{1 + \frac{3}{2} \tan^2 \beta} \left( \because 2tan\alpha = 3tan\beta \right)$
$= \frac{\frac{1}{2} \times tan\beta}{1 + \frac{3}{2} \tan^2 \beta} = \frac{tan\beta}{2 + 3 \tan^2 \beta}$
$= \frac{\frac{sin\beta}{cos\beta}}{2 + 3\frac{\sin^2 \beta}{\cos^2 \beta}} = \frac{\frac{sin\beta}{cos\beta} \times \cos^2 \beta}{2 \cos^2 \beta + 3 \sin^2 \beta}$
$= \frac{sin\beta \times cos\beta}{2 \cos^2 \beta + 2 \sin^2 \beta + \sin^2 \beta}$
$= \frac{1}{2}\frac{2sin\beta \times cos\beta}{2\left( \cos^2 \beta + \sin^2 \beta \right) + \sin^2 \beta}$
$= \frac{1}{2}\frac{\sin2\beta}{\left( 2 + \sin^2 \beta \right)} = \frac{\sin2\beta}{4 + 2 \sin^2 \beta}$
$= \frac{\sin2\beta}{4 + 2\left( 1 - \cos^2 \beta \right)} = \frac{\sin2\beta}{6 - 2 \cos^2 \beta}$
$= \frac{\sin2\beta}{6 - \left( 1 + \cos2\beta \right)} \left( \because 1 + \cos2\beta = 2 \cos^2 \beta \right)$
$= \frac{\sin2\beta}{5 - \cos2\beta} = RHS$
$\text{ Hence proved } .$

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 37 | Page 29