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Evaluate $\csc^{-1} (\sec 5)$ without using a calculator.

I have no idea where to begin on this problem, I've looked over trig identities, and cannot find one that I think applies. Any help would be very much appreciated.

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  • $\begingroup$ Hint: Lets not forget what cosecant (csc) means. cosecant($\theta$) is the secant of the compliment of $\theta$, where two complimentary angles add up to one quarter rotation ($90^\circ$ or $\pi/2$ rad.). $\endgroup$ – John Joy Sep 10 '14 at 14:38
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Hint: Let $\alpha=\csc^{-1}\left(\sec 5\right).$ In other words, $\alpha$ is an angle in the first or fourth quadrant, and $\csc\alpha=\sec 5.$ Put another way, $\sin\alpha=\cos 5.$ Do you see why this is the same thing? Can you take it from here, perhaps by drawing a right triangle?

Note: You will be able to find an exact value this way. It will not be a numerical answer.

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$$ \csc^{-1}\sec5 = \theta $$ $$ \sec 5 = \csc\theta $$ $$ \frac{\text{hypotenuse}}{\text{side adjacent to 5}} = \sec 5 = \csc\theta = \frac{\text{hypotenuse}}{\text{side opposite }\theta} $$ $$ \text{Hence: }~~~~~\text{side adjacent to 5} = \text{side opposite }\theta $$ So $5$ and $\theta$ are the two non-right angles in a right triangle. They add up to a right angle: $$ \theta+5=\text{right angle} $$ (If this is supposed to be in degrees, then $\theta=85$. If it's in radians, we would need the angle between $-\pi/2$ and $+\pi/2$ that has the right cosecant.)

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