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The value of $si{n^6}\theta + co{s^6}\theta + 3si{n^2}\theta .co{s^2}\theta \;$is :-${\mathbf{A}}.\,0 \\ {\mathbf{B}}.\, - 1 \\ {\mathbf{C}}.{\text{ }}1 \\ {\mathbf{D}}.\;{\text{ }}2 \\$

Last updated date: 23rd Mar 2023
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Hint – 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 – 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.