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anushkaudeshan

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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    \[\csc(x)=2\iff \sin(x)=\frac{1}{2}\] 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

  2. anonymous
    • 4 years ago
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    i can work it out if you like.

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

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