Let cos(\alpha +\beta )=\frac{4}{5}  and let sin(\alpha -\beta )=\frac{5}{13} , where 0\leq \alpha , \; \beta \leq \frac{\pi }{4}.  Then tan2\alpha =

  • Option 1)

    \frac{25}{16}

  • Option 2)

    \frac{56}{33}

  • Option 3)

    \frac{19}{12}

  • Option 4)

    \frac{20}{17}

 

Answers (1)
V Vakul

As we leant in

Double Angle Formula -

Double angle formula

- wherein

These are formulae for double angles.

 

 \\ \cos (\alpha + \beta)= \frac{4}{5} \Rightarrow \alpha + \beta = \cos^{-1}\frac{4}{5} ...................(i)

\sin (\alpha - \beta)= \frac{5}{13} \Rightarrow \alpha - \beta = \sin^{-1}\frac{5}{13} ...............(ii)

Adding (1) and (2) 

2 \alpha =cos^{-1}\frac{4}{5}+sin^{-1}\frac{5}{13}

tan \ 2 \alpha =\frac{\tan (\cos^{-1}\frac{4}{5} + \tan (\sin^{-1}\frac{5}{13})}{1- \tan \cos^{1}\frac{4}{5} \tan \sin ^{-1}\frac{3}{13}}

= \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}} = \frac{56}{33}

 


Option 1)

\frac{25}{16}

This option is incorrect.

Option 2)

\frac{56}{33}

This option is correct.

Option 3)

\frac{19}{12}

This option is incorrect.

Option 4)

\frac{20}{17}

This option is incorrect

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions