R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: Exercise 13B

If $\mathrm{acos\theta }+\mathrm{bsin\theta }=\mathrm{m}$ and $\mathrm{asin\theta }-\mathrm{bcos\theta }=\mathrm{n}$, prove that  $\left({\mathrm{m}}^2+{\mathrm{n}}^2\right)=\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)$

If $\mathrm{x}={\mathrm{acos}}^3\mathrm{\theta }$ and $\mathrm{y}={\mathrm{bsin}}^3\mathrm{\theta }$ prove that ${\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{\frac{2}{3}}+{\left(\frac{\mathrm{y}}{\mathrm{b}}\right)}^{\frac{2}{3}}=1$

If $(\mathrm{cot\theta }+\mathrm{tan\theta })=\mathrm{m}$ and  $(\mathrm{sec\theta }-\mathrm{cos\theta })=\mathrm{n}$ prove that ${\left({\mathrm{m}}^2\mathrm{n}\right)}^{\frac{2}{3}}-{\left({\mathrm{mn}}^2\right)}^{\frac{2}{3}}=1$

If $\sec \theta +\tan \theta =\mathrm{p}$, prove that

$\sec \theta =\frac{1}{2}\left(p+\frac{1}{p}\right)$

If $\sec \theta +\tan \theta =p$, Prove that

$\tan \theta =\frac{1}{2}\left(p-\frac{1}{p}\right)$

If $\sec \theta +\tan \theta =p$, Prove that

$\sin \theta =\frac{p^2-1}{p^2+1}$.

If $\tan A=n\tan B$ and $\sin A=m\sin B$, prove that ${\cos }^2\mathrm{A}=\frac{\left({\mathrm{m}}^2-1\right)}{\left({\mathrm{n}}^2-1\right)}$

If $\mathrm{m}=\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)$ and $\mathrm{n}=\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)$ then show that $\sqrt{\frac{\mathrm{m}}{\mathrm{n}}}+\sqrt{\frac{\mathrm{n}}{\mathrm{m}}}=\frac{2}{\sqrt{1-{\tan }^2\mathrm{\theta }}}$.