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November 28, 2014

November 28, 2014

Posted by **David M.** on Wednesday, May 12, 2010 at 11:37pm.

1. sec^sq(tan^-1x)

2. sin(arccot x/(sqrt)1-x^2)

- Trigonometry -
**drwls**, Thursday, May 13, 2010 at 6:57am1. If you mean sec^2[tan^-1x], use the fact that

sec (tan^-1x)= sqrt(x^2+1)/sqrt(1).

Imagine a right triangle with sides x, 1 and sqrt(1+x^2). Side x is opposite the angle with tangent equal to x.

Square the secant and you get x^2 + 1

2. For the angle in question, as part of a right triangle:

adjacent side (not hypotentuse) = x

opposite side = sqrt(1-x^2)

hypotenuse = 1

sin = sqrt(1-x^2)

The last line is your answer.

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