To prove \( (\sin \alpha +\cos \alpha)(\tan \alpha + \cot \alpha)=\sec \alpha +\text{cosec} \alpha \)

Solution:

L.H.S \(= (\sin \alpha +\cos \alpha)(\tan \alpha + \cot \alpha)\)

\(= (\sin \alpha +\cos \alpha)(\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha})\)

\(= (\sin \alpha +\cos \alpha)(\frac{\sin^2 \alpha+\cos^2 \alpha}{\cos \alpha \sin \alpha})\)

\(= (\sin \alpha +\cos \alpha)(\frac{1}{\cos \alpha \sin \alpha})\)       [\(\because \sin^2 \theta + \cos^2 \theta =1\)]

\(= \frac{(\sin \alpha +\cos \alpha)}{\cos \alpha \sin \alpha}\)

\(= \frac{\sin \alpha }{\cos \alpha \sin \alpha}+\frac{\cos \alpha }{\cos \alpha \sin \alpha}\)

\(= \frac{1 }{\cos \alpha}+\frac{1}{ \sin \alpha}\)

\(= \sec \alpha +\text{cosec} \alpha\)

Therefore, \( (\sin \alpha +\cos \alpha)(\tan \alpha + \cot \alpha)=\sec \alpha +\text{cosec} \alpha \)   proved.