question_answer

Prove that: \[(sin\,\theta +cos\,\theta +1).(sin\,\theta -1+cos\,\theta ).sec\,\theta \text{ }.\text{ }cosec\text{ }\theta =2\].

Answer:

L.H.S. \[=(sin\,\theta +cos\,\theta +1).(sin\,\theta -1+cos\,\theta ).sec\,\theta \text{ }.\text{ }cosec\text{ }\theta \]

\[=[(\sin \theta +\cos \theta )+1].[(\sin \theta +\cos \theta )-1].\sec \theta .\cos ec\theta \]

\[=[{{(\sin \theta +\cos \theta )}^{2}}-{{(1)}^{2}}]\sec \theta \,\,\cos ec\theta \]

\[[\because \,\,(a+b)(a-b)={{a}^{2}}-{{b}^{2}}]\]

\[=(2\,\,\sin \theta \cos \theta ).\frac{1}{\cos \theta }.\frac{1}{\sin \theta }\]

\[=2=R.H.S.\]                           Hence Proved.