# Question #b2d25

##### 1 Answer

Mar 28, 2017

# (1-cot^2x)/(cotx-sec^2x) =( 2cos2x ) / ( 2 - sin2x ) #

#### Explanation:

# (1-cot^2x)/(cotx-sec^2x) =(1-(cosx/sinx)^2)/((cosx/sinx)-(1/sinx)^2) #

# " " = (1 - cos^2x/sin^2x ) / ( cosx/sinx - 1/sin^2x ) #

# " " = ( (sin^2x - cos^2x) / sin^2x ) / ( ( cosxsinx - 1)/sin^2x) #

# " " = ( sin^2x - cos^2x ) / ( cosxsinx - 1 ) #

# " " = 2/2 * ( cos^2x -sin^2x) / ( 1 - cosxsinx ) #

# " " = 2 * ( cos^2x -sin^2x) / ( 2 - 2cosxsinx ) #

# " " = 2 * ( cos2x ) / ( 2 - sin2x ) #

That is about as simple as I can get the expression