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anonymous

  • 5 years ago

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  1. anonymous
    • 5 years ago
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    \[(1 + \tan \theta \tan \beta)^{2} + (\tan \theta - \tan \beta)^{2}\] \[= \sec ^{2} \theta \sec ^{2} \beta\]

  2. anonymous
    • 5 years ago
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    Is the question correct, plz recheck it i got some other answer ?

  3. anonymous
    • 5 years ago
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    Yes, its correct. You have to prove that the LHS is equal to the RHS.

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

  6. asnaseer
    • 5 years ago
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    sorry, that last relation should be:\[\tan^2(\alpha)=\sec^2(\alpha)-1\]

  7. anonymous
    • 5 years ago
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    Thanks nikhil and asnaseer for the help.

  8. asnaseer
    • 5 years ago
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    yw

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