Question

The solution of $\sec \theta -cosec\theta =\frac{4}{3},0<\theta <\frac{\pi }{2}$ and $\tan \theta >1$

• A. ${\cos }^{-1}\frac{\sqrt{7}-1}{4}$
• B. ${\sin }^{-1}\frac{\sqrt{7}+1}{4}$
• C. ${\cos }^{-1}\frac{\sqrt{7}+1}{4}$
• D. ${\sin }^{-1}\frac{\sqrt{7}-1}{4}$

Answer 1

Answer: (C), (D)

The correct options are
C ${\cos }^{-1}\frac{\sqrt{7}+1}{4}$
D ${\sin }^{-1}\frac{\sqrt{7}-1}{4}$

$\sec \theta -\csc \theta =\frac{4}{3}$ and $0<\theta <\frac{\pi }{2}$; $\tan \theta >1$
$3(\sin \theta -\cos \theta )=4\sin \theta \cos \theta$
Squaring on both sides,
$9(1-2\sin \theta \cos \theta )=16{\sin }^2\theta {\cos }^2\theta$
$4{\sin }^{2}2\theta +9\sin 2\theta -9=0$
$(4\sin 2\theta -3)(\sin 2\theta +3)=0$
$\sin 2\theta =\frac{3}{4};\sin 2\theta \ne -3$
$\frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{3}{4}$
Simplifying the quadratic equation gives us
$\tan \theta =\frac{4\pm \sqrt{7}}{3}$
But $\tan \theta =\frac{4+\sqrt{7}}{3}$ $(\because \tan \theta >1)$
$\Rightarrow \cos \theta =\frac{\sqrt{7}+1}{4}$ and $\sin \theta =\frac{\sqrt{7}-1}{4}$