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\[

{\mathbf{A}}.\,0 \\

{\mathbf{B}}.\, - 1 \\

{\mathbf{C}}.{\text{ }}1 \\

{\mathbf{D}}.\;{\text{ }}2 \\

\]

Answer
Verified

Hint â€“ Use the formula of \[{(a + b)^3}\] putting \[a = {\sin ^2}\theta \,\,\& \,\,b = {\cos ^2}\theta \].

We know \[{(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\,\,\,\,\,\,\,\,\,......(i)\]

Let \[a = {\sin ^2}\theta \,\,\& \,\,b = {\cos ^2}\theta \]

We also know \[{\sin ^2}\theta + {\cos ^2}\theta = 1\,\,\,\,\,\,\,...........(ii)\]

Then putting the value of \[a\& b\] in \[{\text{(}}i{\text{)}}\]

We get,

\[{({\sin ^2}\theta + {\cos ^2}\theta )^3} = {\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta ({\sin ^2}\theta + {\cos ^2}\theta )\]

\[{\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta (1) = 1\,\,\,\,\,\,\,({\text{from (}}ii{\text{)}})\]

\[{\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta = 1\,\,\,\,\,\,\,({\text{from (}}ii{\text{)}})\]

Hence the answer is \[C\].

Note â€“ In these type of questions of trigonometry we have to use basic concepts like \[{\text{(}}{\sin ^2}\theta + {\cos ^2}\theta = 1)\] and we also used the general formula of \[{{\text{(a + b)}}^{3\;}}\], then we can

get the solution by solving the equation.

We know \[{(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\,\,\,\,\,\,\,\,\,......(i)\]

Let \[a = {\sin ^2}\theta \,\,\& \,\,b = {\cos ^2}\theta \]

We also know \[{\sin ^2}\theta + {\cos ^2}\theta = 1\,\,\,\,\,\,\,...........(ii)\]

Then putting the value of \[a\& b\] in \[{\text{(}}i{\text{)}}\]

We get,

\[{({\sin ^2}\theta + {\cos ^2}\theta )^3} = {\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta ({\sin ^2}\theta + {\cos ^2}\theta )\]

\[{\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta (1) = 1\,\,\,\,\,\,\,({\text{from (}}ii{\text{)}})\]

\[{\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta = 1\,\,\,\,\,\,\,({\text{from (}}ii{\text{)}})\]

Hence the answer is \[C\].

Note â€“ In these type of questions of trigonometry we have to use basic concepts like \[{\text{(}}{\sin ^2}\theta + {\cos ^2}\theta = 1)\] and we also used the general formula of \[{{\text{(a + b)}}^{3\;}}\], then we can

get the solution by solving the equation.

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