Browse by Stream
QnA
Home
Search for Exam, Articles, Questions
QnA
get app
Go To Your Study Dashboard
  1. Home
  2. Solve it, - Trigonometry - JEE Main-3
# Engineering

If \alpha = \cos ^{-1}\left ( \frac{3}{5} \right ),\beta = \tan ^{-1}\left ( \frac{1}{3} \right ), where 0<\alpha ,\beta <\frac{\pi}{2}, then \alpha -\beta is equal to :
 

  • Option 1)

    \sin ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )
     

  • Option 2)

    \tan ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )

  • Option 3)

    \cos ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )

     

  • Option 4)

    \tan ^{-1}\left ( \frac{9}{14} \right )

 

Answers (1)
S solutionqc

\alpha=\cos^{-1}\frac{3}{5}

\beta =\tan^{-1}\left ( \frac{1}{3} \right )\; \; \; 0<\alpha ,\beta <\frac{\pi}{2}

Then

\cos\: \alpha=\frac{3}{5}, \tan \alpha =\frac{4}{3},\alpha =\tan ^{-1}\left ( \frac{4}{3} \right ),

\tan \beta =\frac{1}{3}                        

\alpha -\beta =\cos^{-1}\left ( \frac{3}{5} \right )-\tan^{-1}\left ( \frac{1}{3} \right )

               =\tan^{-1} \frac{4}{3}-\tan^{-1} \frac{1}{3}=\tan^{-1}\left ( \frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{3}\times\frac{1}{3}} \right )

               

               =\tan^{-1}\left ( \frac{1}{1+\frac{4}{9}}\right )=\tan^{-1}\left ( \frac{9}{13} \right )

               \tan^{-1}\left ( \frac{9}{13} \right )

No option matched.


Option 1)

\sin ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )
 

Option 2)

\tan ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )

Option 3)

\cos ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )

 

Option 4)

\tan ^{-1}\left ( \frac{9}{14} \right )

Similar Questions

Preparation Products

Knockout JEE Main April 2021 (One Month)

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

₹ 14000/- ₹ 6999/-
Buy Now
Knockout JEE Main April 2021

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

₹ 22999/- ₹ 14999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Knockout JEE Main April 2022

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

₹ 34999/- ₹ 24999/-
Buy Now
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
Boost your Preparation for JEE Main 2021 with Personlized Coaching
Start Now
 
Exams
Articles
Questions