A) \[{{\sin }^{4}}\theta -{{\cos }^{4}}\theta \]
B) \[1-{{\sin }^{2}}\theta \,\,{{\cos }^{2}}\theta \]
C) \[1+{{\sin }^{2}}\theta \,\,{{\cos }^{2}}\theta \]
D) \[1-3{{\sin }^{2}}\theta \,\,{{\cos }^{2}}\theta \]
Correct Answer: B
Solution :
\[\frac{{{\sin }^{6}}\theta -{{\cos }^{6}}\theta }{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }=\frac{{{(si{{n}^{2}}\theta )}^{3}}-{{({{\cos }^{2}}\theta )}^{3}}}{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }\] \[=\frac{({{\sin }^{2}}\theta -{{\cos }^{2}}\theta )({{\sin }^{4}}\theta +{{\cos }^{4}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta )}{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }\] \[[\because \,\,{{a}^{3}}-{{b}^{3}}=(a-b)\,\,({{a}^{2}}+{{b}^{2}}+ab)]\] \[=\,\,{{\sin }^{4}}\theta +{{\cos }^{4}}\theta +{{\sin }^{2}}\theta .\,co{{s}^{2}}\theta \] \[=\,\,{{({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}^{2}}-2{{\sin }^{2}}\theta .{{\cos }^{2}}\theta +{{\sin }^{2}}\theta .{{\cos }^{2}}\theta \] \[=\,\,\,{{({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}^{2}}-{{\sin }^{2}}\theta .{{\cos }^{2}}\theta \] \[=\,\mathbf{1}-\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}\theta \mathbf{.co}{{\mathbf{s}}^{\mathbf{2}}}\theta \]
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