Question

$\sqrt{\frac{\mathrm{sec\theta }-1}{\mathrm{sec\theta }+1}}+\sqrt{\frac{\mathrm{sec\theta }+1}{\mathrm{sec\theta }-1}}=?$
(a) 2 sin θ
(b) 2 cos θ
(c) 2 cosec θ
(d) 2 sec θ

Answer 1

(c) 2 cosec θ

$$\begin{gathered} \sqrt{\frac{\text{sec}\theta -1}{\text{sec}\theta \text{+1}}}+\sqrt{\frac{\text{sec}\theta \text{+1}}{\text{sec}\theta -1}} \\ \mathrm{By}\mathrm{rationalising},\mathrm{we}\mathrm{get}: \\ \sqrt{\frac{(\text{sec}\theta -1)(\text{sec}\theta -\text{1})}{\left(\text{sec}\theta \text{+1}\right)(\text{sec}\theta -1)}}+\sqrt{\frac{\left(\text{sec}\theta \text{+1}\right)(\text{sec}\theta +1)}{(\text{sec}\theta -1)(\text{sec}\theta +1)}} \\ \text{=}\sqrt{\frac{(\text{sec}\theta -1)(\text{sec}\theta -\text{1})}{({\text{sec}}^2\theta -\text{1})}}+\sqrt{\frac{\left(\text{sec}\theta \text{+1}\right)(\text{sec}\theta +1)}{({\text{sec}}^2\theta -1)}} \\ \text{=}\frac{\sqrt{(\text{sec}\theta -1{)}^2}}{\sqrt{{\text{tan}}^2\theta }}+\frac{\sqrt{(\text{sec}\theta \text{+1}{)}^2}}{\sqrt{{\text{tan}}^2\theta }} \\ \text{=}\frac{(\text{sec}\theta -1)}{\text{tan}\theta }\text{+}\frac{\left(\text{sec}\theta \text{+1}\right)}{\text{tan}\theta } \\ \text{=}\frac{\text{sec}\theta -1+\text{sec}\theta \text{+1}}{\text{tan}\theta } \\ \text{=}\frac{2\text{sec}\theta }{\text{tan}\theta } \\ \text{=}\frac{2\frac{1}{\text{cos}\theta }}{\frac{\text{sin}\theta }{\text{cos}\theta }} \\ \text{=2cosec}\theta \end{gathered}$$