question_answer

Prove the following identity:

\[\frac{{{\sin }^{3}}\theta +{{\cos }^{2}}\theta }{\sin \theta +cos\theta }=1-\sin \theta .\cos \theta \].

Answer:

\[L.H.S.=\frac{{{\sin }^{3}}\theta +{{\cos }^{3}}\theta }{\sin \theta +\cos \theta }\]

\[=\frac{(sin\theta +cos\theta )(si{{n}^{2}}\theta +co{{s}^{2}}\theta -sin\theta .cos\theta )}{(sin\theta +cos\theta )}\]

\[[{{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}+{{b}^{2}}-ab)]\]

\[=1-\sin \theta .\cos \theta =R.H.S.\]

\[[\because \,\,si{{n}^{2}}\theta +co{{s}^{2}}\theta =1]\]              Hence Proved.