Question

Solve this :


$\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathbf{}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\left({\displaystyle \frac{\mathbf{5}\mathbf{\pi }}{\mathbf{2}}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{x}\right)\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\left(}\mathbf{4}\mathbf{\pi }\mathbf{\right)}\mathbf{}\mathbf{sec}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{}\left({\displaystyle \frac{\mathbf{\pi }}{\mathbf{2}}}\right)\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\left({\displaystyle \frac{\mathbf{7}\mathbf{\pi }}{\mathbf{3}}}\right)\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\left({\displaystyle \frac{\mathbf{\pi }}{\mathbf{2}}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\right)}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{3}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathit{x}$

Answer 1

Dear student

$$\begin{gathered} \frac{\tan \left({\displaystyle \frac{5\mathrm{\pi }}{2}}+\mathrm{x}\right)\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }}{\mathrm{cosec}\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)\cot \left({\displaystyle \frac{7\mathrm{\pi }}{3}}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{x}\right)}=\sqrt{3}\mathrm{cosecx} \\ \mathrm{Consider},\mathrm{LHS} \\ \frac{\tan \left({\displaystyle \frac{5\mathrm{\pi }}{2}+\mathrm{x}}\right)\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }}{\mathrm{cosec}\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)\cot \left({\displaystyle \frac{7\mathrm{\pi }}{3}}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}-\mathrm{x}}\right)} \\ =\frac{{\displaystyle \frac{\sin \left({\displaystyle \frac{5\mathrm{\pi }}{2}}+\mathrm{x}\right)}{\cos \left(\frac{5\mathrm{\pi }}{2}+\mathrm{x}\right)}}\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }}{\mathrm{cosec}\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)\cot \left({\displaystyle \frac{7\mathrm{\pi }}{3}}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}-\mathrm{x}}\right)} \\ =\frac{\left({\displaystyle \frac{\cos \left(\frac{5\mathrm{\pi }}{2}\right)\mathrm{sinx}+\mathrm{cosxsin}\left(\frac{5\mathrm{\pi }}{2}\right)}{\cos \left(\frac{5\mathrm{\pi }}{2}\right)\mathrm{cosx}-\sin \left(\frac{5\mathrm{\pi }}{2}\right)\mathrm{sinx}}}\right)\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }}{\mathrm{cosec}\left(\frac{\mathrm{\pi }}{2}\right)\cot \left(\frac{7\mathrm{\pi }}{3}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{x}\right)} \\ =\frac{\left({\displaystyle \frac{-\mathrm{cosx}}{\mathrm{sinx}}}\right)\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }}{\mathrm{cosec}\left(\frac{\mathrm{\pi }}{2}\right)\cot \left(\frac{7\mathrm{\pi }}{3}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{x}\right)}\left[\mathrm{as}\cos \left(\frac{5\mathrm{\pi }}{2}\right)=\cos \left(\frac{\mathrm{\pi }}{2}\right)\mathrm{and}\sin \left(\frac{5\mathrm{\pi }}{2}\right)=1\right] \\ =\frac{\left({\displaystyle \frac{-\mathrm{cosx}}{\mathrm{sinx}}}\right)\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }}{\mathrm{cosec}\left(\frac{\mathrm{\pi }}{2}\right)\cot \left(\frac{7\mathrm{\pi }}{3}\right)\mathrm{cosx}}\left[\mathrm{as}\sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=\mathrm{cosx}\right] \\ =\frac{\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }\left(-\mathrm{cotx}\right)}{\mathrm{cosec}\left(\frac{\mathrm{\pi }}{2}\right)\cot \left(\frac{7\mathrm{\pi }}{3}\right)\mathrm{cosx}} \\ =\frac{-\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }\mathrm{cotx}}{\mathrm{cosec}\left(\frac{\mathrm{\pi }}{2}\right)\cot \left(\frac{7\mathrm{\pi }}{3}\right)\mathrm{cosx}} \\ =\frac{-\cos \left(4\mathrm{\pi }\right)\mathrm{sec\pi }\mathrm{cotx}}{{\displaystyle \frac{\sqrt{3}}{3}}\mathrm{cosx}}\left[\mathrm{as}\mathrm{cosec}\left(\frac{\mathrm{\pi }}{2}\right)=1\mathrm{and}\cot \left(\frac{7\mathrm{\pi }}{3}\right)=\frac{\sqrt{3}}{3}\right] \\ =\frac{-\left(1\right)\mathrm{sec\pi }\mathrm{cotx}}{{\displaystyle \frac{\sqrt{3}}{3}\mathrm{cosx}}}\left[\mathrm{as}\cos 4\mathrm{\pi }=1\mathrm{and}\mathrm{sec\pi }=-1\right] \\ =\frac{\mathrm{cotx}}{\frac{\sqrt{3}}{3}\mathrm{cosx}} \\ =\sqrt{3}\mathrm{cosecx} \\ \mathrm{Identity}\mathrm{used}: \\ \cos (\mathrm{a}+\mathrm{b})=\mathrm{cosacosb}-\mathrm{sinasinb} \\ \sin \left(\mathrm{a}+\mathrm{b}\right)=\mathrm{sinacosb}+\mathrm{cosasinb} \end{gathered}$$
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