Question

Find x from the following equations:
(i) cosec (90° + θ) + x cos θ cot (90° + θ) = sin (90° + θ)
(ii) x cot (90° + θ) + tan (90° + θ) sin θ + cosec (90° + θ) = 0

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}: \\ \mathrm{cosec}\left(90°+\mathrm{\theta }\right)+x\cos \mathrm{\theta }\cot \left(90°+\mathrm{\theta }\right)=\sin \left(90°+\mathrm{\theta }\right) \\ \Rightarrow \sec \mathrm{\theta }+\mathit{}x\mathit{}\cos \mathrm{\theta }\left[-\tan \mathrm{\theta }\right]=\cos \mathrm{\theta } \\ \Rightarrow \sec \mathrm{\theta }-x\mathrm{cos\theta }\mathrm{tan\theta }=\cos \mathrm{\theta } \\ \Rightarrow \sec \mathrm{\theta }-x\mathrm{cos\theta }\times \frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}=\cos \mathrm{\theta } \\ \Rightarrow \sec \mathrm{\theta }-x\sin \mathrm{\theta }=\cos \mathrm{\theta } \\ \Rightarrow \sec \mathrm{\theta }-\cos \mathrm{\theta }=x\sin \mathrm{\theta } \\ \Rightarrow \frac{1}{\cos \mathrm{\theta }}-\mathrm{cos\theta }=x\sin \mathrm{\theta } \\ \Rightarrow \frac{1-{\cos }^2\mathrm{\theta }}{\cos \mathrm{\theta }}=\mathit{}x\sin \mathrm{\theta } \\ \Rightarrow \frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }}=x\sin \mathrm{\theta } \\ \Rightarrow \frac{{\sin }^2\mathrm{\theta }}{\cos \mathrm{\theta }\sin \mathrm{\theta }}=x \\ \Rightarrow \frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}=x \\ \Rightarrow \tan \mathrm{\theta }=x \\ \therefore x=\tan \mathrm{\theta } \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}: \\ x\mathit{}\cot \left(90°+\mathrm{\theta }\right)+\tan \left(90°+\mathrm{\theta }\right)\sin \mathrm{\theta }+\mathrm{cosec}\left(90°+\mathrm{\theta }\right)=0 \\ \Rightarrow x\left[-\tan \mathrm{\theta }\right]+\left[-\cot \mathrm{\theta }\right]\sin \mathrm{\theta }+\sec \mathrm{\theta }=0 \\ \Rightarrow -x\tan \mathrm{\theta }-\cot \mathrm{\theta }\sin \mathrm{\theta }+\sec \mathrm{\theta }=0 \\ \Rightarrow -\mathit{}x\times \frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}-\frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}\times \sin \mathrm{\theta }+\frac{1}{\cos \mathrm{\theta }}=0 \\ \Rightarrow -\mathit{}x\times \frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}-\cos \mathrm{\theta }+\frac{1}{\cos \mathrm{\theta }}=0 \\ \Rightarrow \frac{-x\mathit{}\sin \mathrm{\theta }-{\cos }^2\mathrm{\theta }+1}{\cos \mathrm{\theta }}=0 \end{gathered}$$
^{$$\begin{gathered} \Rightarrow -x\sin \mathrm{\theta }-{\cos }^2\mathrm{\theta }+1=0 \\ \Rightarrow -x\sin \mathrm{\theta }+{\sin }^2\mathrm{\theta }=0 \\ \Rightarrow x\sin \mathrm{\theta }={\sin }^2\mathrm{\theta } \\ \Rightarrow x=\frac{{\sin }^2\mathrm{\theta }}{\sin \mathrm{\theta }} \\ \Rightarrow x=\sin \mathrm{\theta } \end{gathered}$$}