Question

If $\text{cosec}\frac{{360}^o}{7}+\text{cosec}\frac{{540}^o}{7}=\text{cosec}\theta$, where $\theta \in (0,{90}^o)then\theta$

• A. is greater than ${30}^o$
• B. is less than ${20}^o$
• C. lies in $({20}^o,{25}^o)$
• D. lies in $({30}^o,{45}^o)$

Answer 1

The correct option is D lies in $({30}^o,{45}^o)$

${360}^0=2\pi$

${540}^0=3\pi$

Given,

$cosec\frac{2\pi }{7}+cosec\left(\frac{3\pi }{7}\right)=cosec\theta$

$cosec\theta =\frac{1}{\sin \theta }$

$\Rightarrow \frac{1}{\sin \left(\frac{2\pi }{7}\right)}+\frac{1}{\sin \left(\frac{3\pi }{7}\right)}=\frac{1}{\sin \theta }$

$\Rightarrow \frac{\sin \left(\frac{3\pi }{7}\right)+\sin \left(\frac{2\pi }{7}\right)}{\sin \left(\frac{2\pi }{7}\right)\sin \left(\frac{3\pi }{7}\right)}=\frac{1}{\sin \theta }$

$\{\because sinC+sinD=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\}$

$\Rightarrow \frac{2\sin \left(\frac{5\pi }{7}\right)\cos \left(\frac{\pi }{7}\right)}{\sin \left(\frac{2\pi }{7}\right)\sin \left(\frac{3\pi }{7}\right)}=\frac{1}{\sin \theta }$

$\Rightarrow \sin \theta =\frac{\sin \left(\frac{2\pi }{7}\right)\sin \left(\frac{3\pi }{7}\right)}{2\sin (\pi -\frac{2\pi }{7})\cos \left(\frac{3\pi }{7}\right)}$

$\Rightarrow \sin \theta =\frac{\sin \left(\frac{2\pi }{7}\right)\sin \left(\frac{3\pi }{7}\right)}{2\sin \left(\frac{2\pi }{7}\right)\cos \left(\frac{\pi }{7}\right)}$

$sin\theta =\frac{\sin \left(\frac{3\pi }{7}\right)}{2\cos \left(\frac{\pi }{7}\right)}=$

$\sin \left(\frac{3\pi }{7}\right)=0.9749$

$\cos \left(\frac{\pi }{7}\right)=0.9009$

$\Rightarrow \sin \theta =0.5410$

${\sin 30}^0=1/2=0.5,{\sin 60}^0=\sqrt{3}/2=0.865$