Question

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

Answer 1

Hint – Use the formula of $$(a+b{)}^3$$ putting $$a={\sin }^2\theta \mathrm{\&}b={\cos }^2\theta$$.

We know $$(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2......(i)$$
Let $$a={\sin }^2\theta \mathrm{\&}b={\cos }^2\theta$$
We also know $${\sin }^2\theta +{\cos }^2\theta =1...........(ii)$$
Then putting the value of $$a\mathrm{\&}b$$ 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.