Question

cosecx = 2, so what is (cot^2x + 3tan^2x)/8sin^2x - cosec^2x + 6

Answer 1

\[\csc(x)=2\iff \sin(x)=\frac{1}{2}\]
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

Answer 2

i can work it out if you like.

Answer 3

\[\left(\left.\left(\text{Cot}[x]^2 + 3 \text{ Tan}[x]^2\right)\right/8 \text{ Sin}[x]^2\right)- \text{Csc}[x]^2+ 6 \]Replace x with the ArcCsc[2] and simplify.\[\frac{17}{8} \]