Question

If 5tan A=$\sqrt{7}$ ,where $\pi <A<\frac{3\pi }{2}$ and sec B=$\sqrt{11}$, where $\frac{3\pi }{2}<B<2\pi$, find the value of cosec A - tan B.

Answer 1

$\tan A=\frac{\sqrt{7}}{5}\sec B=\sqrt{11}$

$\cot A=\frac{5}{\sqrt{7}}$

$\therefore 1+{\cos }^{2}A={\csc }^{2}A$

$\Rightarrow {\csc }^{2}A=1+{\left(\frac{5}{\sqrt{7}}\right)}^2$

$=1+\frac{25}{7}$

${\csc }^{2}A=\frac{32}{7}$

$\csc A=\pm \sqrt{\frac{32}{7}}=\sqrt{\frac{32}{7}}$

Now,

$1+{\tan }^{2}B={\sec }^{2}B$

$\Rightarrow {\tan }^{2}B={\sec }^{2}B-1=11-1$

$\tan B=\sqrt{10}=-\sqrt{10}$

$\therefore \csc A-\tan B=-\sqrt{\frac{32}{7}}-(-\sqrt{10})$

$\csc A-\tan B=\sqrt{10}-\sqrt{\frac{32}{7}}$