logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
0 votes

If $\cos(\theta+\phi)=m\cos(\theta-\phi)$ then prove that $\tan\theta=\large\frac{1-m}{1+m}$$\cot\phi$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\cos a+\cos b=2\cos(\large\frac{a+b}{2})$$\cos(\large\frac{a-b}{2})$
  • $\cos a-\cos b=-2\sin(\large\frac{a+b}{2})$$\sin (\large\frac{a-b}{2})$
Given
$\cos(\theta+\phi)=m\cos(\theta-\phi)$
$\large\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}=\frac{m}{1}$
Applying componendo and dividendo
$\large\frac{cos(\theta+\phi)+\cos(\theta-\phi)}{cos(\theta+\phi)-\cos(\theta-\phi)}=\frac{m+1}{m-1}$
$\large\frac{2\cos(\Large\frac{\theta+\phi+\theta-\phi)}{2}.\cos(\Large\frac{\theta+\phi-\theta+\phi}{2})}{-2\sin(\Large\frac{\theta+\phi+\theta-\phi)}{2}.\sin(\Large\frac{\theta+\phi-\theta+\phi}{2})}=\frac{m+1}{m-1}$
$\large\frac{\cos \theta.\cos\phi}{-\sin \theta.\sin \theta}=\frac{m+1}{m-1}$
$-\cot \theta.\cot\phi=\large\frac{m+1}{m-1}$
$-\cot \theta=\large\frac{m+1}{m-1}\frac{1}{\cot \phi}$
$-\large\frac{1}{\tan \theta}=\frac{m+1}{m-1}$$\times \large\frac{1}{\cot \phi}$
$-\cot \phi.\large\frac{m-1}{1+m}=$$\tan \theta$
$\tan\theta=\large\frac{1-m}{1+m}$$\cot\phi$
Hence proved.
answered Apr 20, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...