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

Prove that $\tan(\large\frac{\pi}{4}$$+\theta)+\tan(\large\frac{\pi}{4}-$$\theta)=2\sec 2\theta$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A.\tan B}$
  • $\cos 2A=\large\frac{1-\tan^2A}{1+\tan^2A}$
$\tan(\large\frac{\pi}{4}$$+\theta)+\tan(\large\frac{\pi}{4}-$$\theta)$
$\large\frac{\tan\Large\frac{\pi}{4}+\tan\theta}{1-\tan \Large\frac{\pi}{4}.\tan\theta}+\frac{\tan\Large\frac{\pi}{4}-\tan \theta}{1+\tan \Large\frac{\pi}{4}.\tan\theta}$
$\Rightarrow \large\frac{1+\tan \theta}{1-\tan \theta}+\large\frac{1-\tan \theta}{1+\tan \theta}$
$\Rightarrow \large\frac{(1+\tan \theta)^2+(1-\tan \theta)^2}{(1-\tan \theta)(1+\tan \theta)}$
$\Rightarrow \large\frac{(1+\tan^2\theta+2\tan \theta)+(1+\tan^2\theta-2\tan \theta)}{1-\tan^2\theta}$
$\Rightarrow \large\frac{2(1+\tan^2\theta)}{1-\tan^2\theta}$
$\Rightarrow \large\frac{2}{\cos 2\theta}$
$\Rightarrow \sec 2\theta$=RHS
Hence proved
answered May 21, 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
...