Question

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

Answer 1

$\begin{array}{c}\tan A=\frac{\sqrt{7}}{5}\\ \Rightarrow \sec B=\frac{\sqrt{11}}{1},{\sec }^{2}B=11,{\sec }^{2}B-1=10\\ \therefore \tan B=\sqrt{10}\\ \Rightarrow \tan A=\frac{\sqrt{7}}{5}\\ squaringonbothside\\ \Rightarrow {\tan }^{2}A=\frac{7}{25},{\cot }^{2}A=\frac{25}{7},It{\cot }^{2}A=\frac{32}{7}\\ \therefore \cos e{c}^{2}A=\frac{32}{7},\cos ecA=\sqrt{\frac{32}{7}}\\ \therefore \cos ecA-\tan B\\ \Rightarrow \sqrt{\frac{32}{7}}-\sqrt{10}\\ \frac{\left(4\sqrt{2}-10\right)}{7}Ans.\end{array}$