Question

cos theta/p = sin theta/ q, then value of p/ sec 2theta + q/ cosec 2theta is
a. p
b. q
c. pq
d. p/q

Answer 1

Hi,

$$\begin{gathered} \mathrm{Let}\frac{\mathrm{cosx}}{\mathrm{p}}=\frac{\mathrm{sinx}}{\mathrm{q}}=\mathrm{k} \\ \mathrm{so}\mathrm{sinx}=\mathrm{kq}\mathrm{and}\mathrm{cosx}=\mathrm{pk} \\ \mathrm{tanx}=\frac{\mathrm{q}}{\mathrm{p}}\mathrm{and}\mathrm{hence}\mathrm{sinx}=\frac{\mathrm{q}}{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}}:\mathrm{cosx}=\frac{\mathrm{p}}{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}} \\ \mathrm{Now}, \\ \frac{\mathrm{p}}{\sec 2\mathrm{x}}+\frac{\mathrm{q}}{\mathrm{cosec}2\mathrm{x}} \\ =\mathrm{pcos}2\mathrm{x}+\mathrm{qsin}2\mathrm{x} \\ =\mathrm{p}\left(2{\cos }^2\mathrm{x}-1\right)+2\mathrm{qsinxcosx} \\ =\mathrm{p}\left(2\times \frac{{\mathrm{p}}^2}{{\mathrm{p}}^2+{\mathrm{q}}^2}-1\right)+\frac{2{\mathrm{pq}}^2}{{\mathrm{p}}^2+{\mathrm{q}}^2} \\ =\frac{\mathrm{p}\left({\mathrm{p}}^2-{\mathrm{q}}^2+2{\mathrm{q}}^2\right)}{{\mathrm{p}}^2+{\mathrm{q}}^2} \\ =\frac{\mathrm{p}\left({\mathrm{p}}^2+{\mathrm{q}}^2\right)}{{\mathrm{p}}^2+{\mathrm{q}}^2}=\mathrm{p} \end{gathered}$$
Regards