Question

Prove:

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

Thanks nikhil and asnaseer for the help.

Answer 7

yw