Question

If (sin θ + cos θ) = p and (sec θ + cosec θ) = q, then q(p² − 1) = ?
(a) 2
(b) 2p
(c) $\frac{1}{p^2}$
(d) $\frac{1}{p}$

Answer 1

(b) 2p

$$\begin{gathered} \text{Given:}\mathrm{sin\theta }+\mathrm{cos\theta }=\mathit{}p \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ {\mathit{\left(}\mathit{\text{sin}}\theta \mathit{\text{+cos}}\theta \mathit{\right)}}^{\mathit{2}}\mathit{=}{p}^{\mathit{2}}\mathit{...}\mathit{(}\mathit{\text{i}}\mathit{)} \\ \text{and}\left(\text{sec}\theta \text{+cosec}\theta \right)=q...\left(ii\right) \\ \mathrm{F}\text{rom (i),}({\text{sin}}^2\theta +{\text{cos}}^2\theta )+2\text{sin}\theta \text{cos}\theta =p^2 \\ \text{=>1+}2\text{sin}\theta \text{cos}\theta =p^2 \\ \text{=>}2\text{sin}\theta \text{cos}\theta =p^2-1 \\ \text{=>sin}\theta \text{cos}\theta =\frac{p^2-1}{2}...\left(\text{iii}\right) \\ \text{According to the question, we have:} \\ \left(\text{sec}\theta \text{+cosec}\theta \right)=q \\ \text{=>}(\frac{1}{\text{cos}\theta }+\frac{1}{\text{sin}\theta })=q \\ \text{=>}\left(\frac{\text{sin}\theta \text{+cos}\theta }{\cos \theta \text{sin}\theta }\right)=q \\ \text{Hence,}\left(\frac{p}{\frac{p^2-1}{2}}\right)\text{=}q \\ \text{=>}\frac{2p}{p^2-1}\text{=}q \\ \text{=>}q(p^2-1)\text{=2}p \end{gathered}$$