Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

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 \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .

 
Advertisement Remove all ads

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
  Is there an error in this question or solution?
Advertisement Remove all ads

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
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×