Posted by **abby** on Sunday, August 28, 2005 at 12:34am.

evaluate the following expressions:

tan(sec^-1(5/3))

tan(sec^-1(25/7))

cot(csc^-1(5/3))

i know the answers.. i jus don't know how to solve them =( PLEASE help me

I assume your sec^-1 notation denot4es the arcsecant function, etc.

sec^-1 (5/3) = cos^-1 (3/5). Think of a 3,4,5 right triangle. The tangent of the angle is 4/3.

sec^-1 (25/7) = cos^-1 (7/25)

Think of a 7,25,24 right triangle. The hypotenuse is 25 and the adjacent side is 7. The tangent is 24/7

csc^(5/3) = sin^-1 (3/5). Think of a 3,4,5 right triangle again. The hypotenuse is 5 and the opposite side of the angle in question is 3. The cotangent is (adjacent side)/(opposite side) = 4/3.

## Answer This Question

## Related Questions

- Integration - Intergrate ¡ì sec^3(x) dx could anybody please check this answer. ...
- more trig.... how fun!!!! - if you can't help me with my first question hopw you...
- trigonometry repost - Reduce (csc^2 x - sec^2 X) to an expression containing ...
- Math - Trigonometry - Verify the following: 1. cos x/(1-sinx)= sec x + tan x 2...
- Pre Calculus. - Can someone check my answers please!!! Simplify (tan ^2 theta ...
- Calculus 12th grade (double check my work please) - 2- given the curve is ...
- calculus (check my work please) - Not sure if it is right, I have check with the...
- trigonometry - Prove: 1) 1 / sec X - tan X = sec X + tan X 2) cot A + tan A = ...
- Math - Trig - I'm trying to verify these trigonometric identities. 1. 1 / [sec(x...
- Precalculus - Circle O below has radius 1. Eight segment lengths are labeled ...

More Related Questions