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.
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