Question

Question 18

$$\begin{gathered} 18.co{t}^{-1}\left[{\left(\cos \alpha \right)}^{1/2}\right]-{\tan }^{-1}\left[{\left(\cos \alpha \right)}^{1/2}\right]=x,then\sin x= \\ \left(A\right){\tan }^2\left(\frac{\alpha }{2}\right)\left(B\right)co{t}^2\left(\frac{\alpha }{2}\right)\left(C\right)\tan \alpha \left(D\right)\tan \left(\frac{\alpha }{2}\right) \end{gathered}$$

Answer 1

Dear student
$$\begin{gathered} \mathrm{x}={\cot }^{-1}\sqrt{\mathrm{cos\alpha }}-{\tan }^{-1}\sqrt{\mathrm{cos\alpha }} \\ \mathrm{x}=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\sqrt{\mathrm{cos\alpha }}-{\tan }^{-1}\sqrt{\mathrm{cos\alpha }} \\ \mathrm{x}=\frac{\mathrm{\pi }}{2}-2{\tan }^{-1}\sqrt{\mathrm{cos\alpha }} \\ \mathrm{x}=\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\left\{\frac{1-{\left(\sqrt{\mathrm{cos\alpha }}\right)}^2}{1+{\left(\sqrt{\mathrm{cos\alpha }}\right)}^2}\right\}\left[\mathrm{As}2{\tan }^{-1}\mathrm{x}={\cos }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right)\right] \\ \mathrm{x}=\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\left(\frac{1-\mathrm{cos\alpha }}{1+\mathrm{cos\alpha }}\right) \\ \mathrm{x}=\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\left({\tan }^2\frac{\mathrm{\alpha }}{2}\right) \\ {\cos }^{-1}\left({\tan }^2\frac{\mathrm{\alpha }}{2}\right)=\frac{\mathrm{\pi }}{2}-\mathrm{x} \\ {\tan }^2\left(\frac{\mathrm{\alpha }}{2}\right)=\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right) \\ {\tan }^2\frac{\mathrm{\alpha }}{2}=\mathrm{sinx} \end{gathered}$$
Regards