Question

if a cos theta + b sin theta is equal to 6 and a sin theta minus B cos theta is equal to 8 then a square + b square is equal to

Answer 1

Dear Student,

Solution:
$$\begin{gathered} \mathrm{acos\theta }+\mathrm{bsin\theta }=6 \\ \mathrm{asin\theta }-\mathrm{bcos\theta }=8 \end{gathered}$$
Now, Squaring and adding these two equations:

$$\begin{gathered} {\mathrm{a}}^2{\cos }^2\mathrm{\theta }+{\mathrm{b}}^2{\sin }^2\mathrm{\theta }+2\mathrm{abcos\theta sin\theta }=36 \\ {\mathrm{a}}^2{\sin }^2\mathrm{\theta }+{\mathrm{b}}^2{\cos }^2\mathrm{\theta }-2\mathrm{abcos\theta sin\theta }=64 \\ \mathrm{Adding}: \\ {\mathrm{a}}^2({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })+{\mathrm{b}}^2({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta })+2\mathrm{abcos\theta sin\theta }-2\mathrm{abcos\theta sin\theta }=36+64 \\ \mathrm{Using}{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1, \\ {\mathrm{a}}^2+{\mathrm{b}}^2=100 \end{gathered}$$


​​​​​​Regards