## anonymous 4 years ago Prove:

1. anonymous

$(1 + \tan \theta \tan \beta)^{2} + (\tan \theta - \tan \beta)^{2}$ $= \sec ^{2} \theta \sec ^{2} \beta$

2. anonymous

Is the question correct, plz recheck it i got some other answer ?

3. anonymous

Yes, its correct. You have to prove that the LHS is equal to the RHS.

4. anonymous

5. asnaseer

\begin{align} (1 + \tan \theta \tan \beta)^{2} + (\tan \theta - \tan \beta)^{2}&=1+\tan^2(\theta)\tan^2(\beta)+2\tan(\theta)\tan(\beta)\\ &+\tan^2(\theta)-2\tan(\theta)\tan(\beta)+\tan^2(\beta)\\ &=1+\tan^2(\theta)+\tan^2(\beta)+\tan^2(\theta)\tan^2(\beta) \end{align}then use the relation:$\sec^2(\alpha)=\tan^2(\alpha)-1$in the above and the answer should "pop out" :)

6. asnaseer

sorry, that last relation should be:$\tan^2(\alpha)=\sec^2(\alpha)-1$

7. anonymous

Thanks nikhil and asnaseer for the help.

8. asnaseer

yw