Thursday

January 29, 2015

January 29, 2015

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

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

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

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

i know the answers.. i jus dont 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...

Math - Trig - I'm trying to verify these trigonometric identities. 1. 1 / [sec(x...

trig - If sinx = -3/5, tan x > 0, sec y = -13/5 and cot y < 0, find the ...

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...

Calculus - could anybody please explain how sec x tan x - ¡ì sec x tan^2(x) dx...

calculus - find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx ...