anonymous
  • anonymous
Prove:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[(1 + \tan \theta \tan \beta)^{2} + (\tan \theta - \tan \beta)^{2}\] \[= \sec ^{2} \theta \sec ^{2} \beta\]
anonymous
  • anonymous
Is the question correct, plz recheck it i got some other answer ?
anonymous
  • anonymous
Yes, its correct. You have to prove that the LHS is equal to the RHS.

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anonymous
  • anonymous
asnaseer
  • 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" :)
asnaseer
  • asnaseer
sorry, that last relation should be:\[\tan^2(\alpha)=\sec^2(\alpha)-1\]
anonymous
  • anonymous
Thanks nikhil and asnaseer for the help.
asnaseer
  • asnaseer
yw

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