Trig functions: 'visual' explanation of sec[arcsin(x-1)]

[allie!!!!!]

I'm a very visual person and I was wondering if someone could step by step show me how to solve this type of problem

sec[arcsin(x-1)]

thanks in advance

 

[skeeter]

o.k. visual person ... draw a right triangle

label one of the acute angles \(\displaystyle \L \theta\)

let \(\displaystyle \L \theta = \arcsin(x-1)\)

this is another way of saying that \(\displaystyle \L \sin\theta = x-1\)

since the sine of an angle is opposite/hypotenuse ... \(\displaystyle \L \sin\theta = \frac{x-1}{1}\)

label the side opposite angle \(\displaystyle \L \theta\) as \(\displaystyle \L x-1\).

label the hypotenuse 1.

using pythagoras, the adjacent side will be \(\displaystyle \L \sqrt{1^2 - (x-1)^2} = \sqrt{2x-x^2}\)

\(\displaystyle \L \sec[\arcsin(x-1)] = \sec\theta = \frac{hypotenuse}{adjacent} = \frac{1}{\sqrt{2x-x^2}}\)