Question

Prove the following :
(viii) $\sqrt{se{c}^2\theta +\cos e{c}^2\theta }=\tan \theta +cot\theta$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sqrt{se{c}^2\theta +\cos e{c}^2\theta } \\ =\sqrt{1+{\tan }^2\theta +1+co{t}^2\theta } \\ (\sin ce,1+{\tan }^2\theta =se{c}^2\theta ,1+co{t}^2\theta =\cos e{c}^2\theta ) \\ =\sqrt{{\tan }^2\theta +co{t}^2\theta +2} \\ =\sqrt{{\tan }^2\theta +co{t}^2\theta +2\tan \theta .cot\theta }(\because \tan \theta .cot\theta =1) \\ =\sqrt{{\left(\tan \theta +cot\theta \right)}^2} \\ =\tan \theta +cot\theta \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$