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  • Prove the following :tan A by 1+ sec A - tan A by 1- sec A = 2cosecA

 Prove the following :

\frac{tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}= 2cosecA  

Answers (1)

Solution.     

 \frac{tan A}{1+\sec A}-\frac{\tan A}{1-\sec A} = 2cosecA
              Taking L.H.S.
      = \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}
Taking L.C.M.
=\frac{\tan A\left ( 1-\sec A \right )-\tan A\left ( 1+\sec A \right )}{\left ( 1+\sec A \right )\left ( 1-\sec A \right )}
=\frac{\tan A-\tan A\\sec A-\tan A-\tan A\sec A}{1-\sec ^{2}A}   \left ( \because \left ( a-b \right )\left ( a+b \right ) = a^{2}-b^{2}\right )
= \frac{-2\tan A\sec A}{-\tan ^{2}A}                 \left ( \because \sec ^{2}A-\tan ^{2}A= 1 \right )
= \frac{2\sec A}{\tan A}
= \frac{2}{\cos A}\times \frac{\cos A}{\sin A}              \begin{pmatrix} \because \sec \theta = \frac{1}{\cos \theta } & \\ \because \tan \theta =\frac{\sin \theta }{\cos \theta } & \end{pmatrix}
= \frac{2}{\sin A}= 2\cos ec A   \left ( \because \frac{1}{\sin \theta }= \cos ec\theta \right )

L.H.S. = R.H.S.

Hence proved

Posted by

infoexpert27

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