Here's the question you clicked on:
anushkaudeshan
cosecx = 2, so what is (cot^2x + 3tan^2x)/8sin^2x - cosec^2x + 6
\[\csc(x)=2\iff \sin(x)=\frac{1}{2}\] gives \[\cos(x)=\pm\frac{\sqrt{3}}{2}\] and so \[\tan^2(x)=(\frac{1}{2}\times \frac{2}{\sqrt{3}})^2=\frac{1}{3}\] and so on for the rest
i can work it out if you like.
\[\left(\left.\left(\text{Cot}[x]^2 + 3 \text{ Tan}[x]^2\right)\right/8 \text{ Sin}[x]^2\right)- \text{Csc}[x]^2+ 6 \]Replace x with the ArcCsc[2] and simplify.\[\frac{17}{8} \]