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2015-04-16T19:45:03+05:30

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\dfrac{\csc^2\alpha-\sec^2\alpha}{\csc^2\alpha+\sec^2\alpha}

Using reciprocal identities \csc\alpha = \frac{1}{\sin\alpha}, ~\sec\alpha=\frac{1}{\cos\alpha} the given expression becomes
\dfrac{\frac{1}{\sin^2\alpha}-\frac{1}{\cos^2\alpha}}{\frac{1}{\sin^2\alpha}+\frac{1}{\cos^2\alpha}}

Multiply top and bottom by \sin^2\alpha and get
\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}

Plugin the given value and simplify
\dfrac{1-\left(\frac{1}{7}\right)^2}{1+\left(\frac{1}{7}\right)^2}~=~\dfrac{49-1}{49+1}~=~\dfrac{48}{50}~=~\dfrac{24}{25}
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hey i asked another question, can u try to do it plzz
:)
i tried a bit and gave up as it got complicated lol il give it a try again..
ok np
;)