Question

If $\alpha \in\left(\cos ^{-1} \frac{\sqrt{5}-1}{2}, \frac{\pi}{2}\right)$ then which one of the following inequalities hold good?

• A. $\cos \alpha<\sin \alpha<\cot \alpha<\tan \alpha$
• B. $\cos \alpha<\cot \alpha<\sin \alpha<\tan \alpha$
• C. $\sin \alpha<\tan \alpha<\cos \alpha<\cot \alpha$
• D. $\sin \alpha<\cot \alpha<\tan \alpha<\cos \alpha$

Answer 1

Answer: (C)

As $\alpha \in\left(\cos ^{-1} \frac{\sqrt{5}-1}{2}, \frac{\pi}{2}\right)$ so $\cos \alpha \in\left(0, \frac{\sqrt{5}-1}{2}\right)$
[Online test-4, P-1]
Now $(\cos \alpha-\cot \alpha)=\cos \alpha\left(\frac{\sin \alpha-1}{\sin \alpha}\right)<0 \Rightarrow \cos \alpha<\cot \alpha$
Also $(\sin \alpha-\tan \alpha)=\sin \alpha\left(\frac{\cos \alpha-1}{\cos \alpha}\right)<0 \Rightarrow \sin \alpha<\tan \alpha$
And $(\cot \alpha-\sin \alpha)=\frac{1}{\sin \alpha}\left(\cos \alpha-\sin ^2 \alpha\right)=\frac{1}{\sin \alpha}\left(\cos \alpha-1+\cos ^2 \alpha\right)=\frac{1}{\sin \alpha}\left(\left(\cos \alpha+\frac{1}{2}\right)^2-\frac{5}{4}\right)$
$=\frac{1}{\sin \alpha}\left(\cos \alpha-\frac{\sqrt{5}-1}{2}\right)\left(\cos \alpha+\frac{\sqrt{5}-1}{2}\right)<0 \forall \cos \alpha \in\left(0, \frac{\sqrt{5}-1}{2}\right)$
$\therefore(\cot \alpha-\sin \alpha)<0 \Rightarrow \cot \alpha<\sin \alpha$
$\therefore$ From inequation (1), (2) and (3), we get $\cos \alpha<\cot \alpha<\sin \alpha<\tan \alpha$.