Question

If $p_1,p_2,p_3$ are the principal values of following trigonometric equations
1)$\sin \theta =-\frac{1}{\sqrt{2}}$
2)$\cos \theta =-\frac{\sqrt{3}}{2}$
3)$\tan \theta =\sqrt{3}-2$

• A. $p_1<p_2<p_3$
• B. $p_1<p_3<p_2$
• C. $p_3<p_1<p_2$
• D. $p_2<p_3<p_1$

Answer 1

The correct option is B $p_1<p_3<p_2$

$\sin \theta =\frac{-1}{\sqrt{2}}\Rightarrow \theta ={\sin }^{-1}\left(\frac{-1}{\sqrt{2}}\right)\Rightarrow \theta =\frac{-\pi }{4}$
as Range of Principal Value of ${\sin }^{-1}$ is $[\frac{-\pi }{2},\frac{\pi }{2}]$

$\cos \theta =-\frac{\sqrt{3}}{2}\Rightarrow \theta ={\cos }^{-1}(-\frac{\sqrt{3}}{2})\Rightarrow \theta =\frac{2\pi }{3}$
as Range of Principal Value of $co{s}^{-1}$ is $[0,\pi ]$

$\tan \theta =\sqrt{3}-2\Rightarrow \theta ={\tan }^{-1}(\sqrt{3}-2)\Rightarrow \theta =\frac{\pi }{2}+\frac{\pi }{12}=\frac{7\pi }{12}$
as Range of Principal Value of ${\tan }^{-1}$ is $(\frac{-\pi }{2},\frac{\pi }{2})$

Since, $\frac{-\pi }{4}<\frac{7\pi }{12}<\frac{2\pi }{3}$

Therefore, $p_1<p_3<p_2$