Question

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

Answer 1

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.