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  • If tanθ + secθ =l, then prove that secθ= l power 2 +1 by 2l

If  tanθ + secθ =\l, then prove that

sec\theta=\frac{\l ^{2}+1}{2\l }

Answers (1)

Solution.         Given : tanθ + secθ =l
There fore
\frac{\l ^{2}+1}{2\l }=\frac{\left ( \tan \theta +\sec \theta \right )^{2}+1}{2\left ( \tan \theta +\sec \theta \right )}
\frac{\tan^{2} \theta +\sec^{2} \theta+2\tan \theta \sec \theta+1}{2\left ( \tan \theta \right )+2\sec \theta }
\frac{\sec^{2} \theta +\sec^{2} \theta+2\tan \theta \sec \theta}{2\left ( \tan \theta+\sec \theta \right ) }       \left [ \because 1+\tan ^{2}\theta = \sec ^{2}\theta \right ]
\frac{2\sec^{2} \theta +2\tan \theta \sec \theta}{2\left ( \tan \theta+2\sec \theta \right ) }
\frac{2\sec \theta \left ( \sec \theta+\tan \theta \right ) }{2\left ( \tan \theta+\sec \theta \right ) }
= \sec \theta (R.H.S)
Hence proved

Posted by

infoexpert27

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