SOLUTION: Please help me solve this equation: (cscX-cscXcos^2X)/(sinXtanX)=cotX

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me solve this equation: (cscX-cscXcos^2X)/(sinXtanX)=cotX       Log On


   

Question 955024: Please help me solve this equation: (cscX-cscXcos^2X)/(sinXtanX)=cotX

Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
(cscX-cscXcos^2X)/(sinXtanX)=cotX
Multiply by sin*tan
csc*(1 - sin^2) = sin
Multiply by sine
1 - sin^2 = sin^2
2sin^2 = 1
sin^2 = 1/2
sin = ħsqrt(2)/2
x = n*45 degs n = 1,2,3...
x = n*pi/4 radians n = 1,2,3...

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
it appears this one can be solve by converting everything to sines and cosines and secants and cosecants.

here's how:

work with the left side of the equation.

the expression on the left side of the equation is :

%28csc%28x%29-csc%28x%29cos%5E2%28x%29%29%2F%28sin%28x%29tan%28x%29%29

you can factor out csc%28x%29 in the numerator to get:

csc%28x%29+%2A+%281+-+cos%5E2%28x%29%29+%2F+%28sin%28x%29+%2A+tan%28x%29%29

since +1+-+cos%5E2%28x%29%29 is equal to sin%5E2%28x%29, your expression becomes:

%28csc%28x%29+%2A+sin%5E2%28x%29%29+%2F+%28sin%28x%29+%2A+tan%28x%29%29

since sin%5E2%28x%29+%2F+sin%28x%29 is equal to sin%28x%29, your expression becomes:

%28csc%28x%29+%2A+sin%28x%29%29+%2F+tan%28x%29

since csc%28x%29+=+1%2Fsin%28x%29 and since sin%28x%29+%2F+sin%28x%29 is equal to 1, your expression becomes:

1+%2F+tan%28x%29.

since +1+%2F+tan%28x%29 is equal to cot%28x%29, then your expression becomes:

cot%28x%29

since the right side of your equation is also equal to cot%28x%29, you are done and the identity has been proven.