29 and or also

Dear Student,
Solution) 
$$\begin{gathered} Given: \sin \theta +\cos \theta =a \\ Squaring on both sides, we get, \\ {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =a^2 \\ \Rightarrow \sin \theta \cos \theta =\frac{a^2-1}{2}...\left(i\right) \{because {\sin }^2\theta +{\cos }^2\theta =1\} \\ Now, \\ {\sin }^6\theta +{\cos }^6\theta ={\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3=\left({\sin }^2\theta +{\cos }^2\theta \right)\left[{\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2-{\sin }^2\theta {\cos }^2\theta \right] \\ ={\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta \\ ={\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-2{\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta \\ =1-3{\sin }^2\theta {\cos }^2\theta \\ =1-3{\left(\frac{a^2-1}{2}\right)}^2 \left\{U\sin g\right(i\left)\right\} \\ =1-\frac{3{\left(a^2-1\right)}^2}{4} \\ =\frac{4-3{\left(a^2-1\right)}^2}{4} \end{gathered}$$
Dear kindly post the second query in another thread.
Regards!