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  • Prove the following : (sin α + cos α) (tan α + cot α) = sec α + cosec α

Prove the following : (sin α + cos α) (tan α + cot α) = sec α + cosec α

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Solution.  
(sin α + cos α) (tan α + cot α) = sec α + cosec α
Taking L.H.S.
\left ( \sin \alpha +\cos \alpha \right )\left ( \tan \alpha +\cot \alpha \right )
= \left ( \sin \alpha +\cos \alpha \right )\left ( \frac{\sin \alpha }{\cos \alpha } +\frac{\cos \alpha}{\sin \alpha } \right )\left ( \because \tan \theta = \frac{\sin \theta }{\cos \theta },\cot \theta = \frac{\cos \theta }{\sin \theta } \right )
Taking L.C.M.
= \frac{\left ( \sin \alpha +\cos \alpha \right )\left ( \sin ^{2}\alpha +\cos^{2} \alpha \right )}{\sin \alpha +\cos \alpha }
= \frac{\sin \alpha }{\sin \alpha \cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }              \left ( \because \sin^{2} \theta +\cos^{2} \theta = 1 \right )
by separately divide
= \frac{\sin \alpha }{\sin \alpha \cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }
= \frac{1}{\cos \alpha }+\frac{1}{\sin \alpha }
= \sec \alpha +\cos ec \, \alpha          \left ( \because \frac{1}{\cos \theta }= \sec \theta ,\frac{1}{\sin \theta }= \cos ec\theta \right )
L.H.S. = R.H.S.
Hence proved.




 

 

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