Question #2dbc6

1 Answer
Jan 21, 2018

#1/(cos(theta)sin(theta)) " or " 2csc2theta #

Explanation:

#cot(theta)+tan(theta)#

Its equal to

#1/tan(theta)+tan(theta)#

Get the lcd and add which will result to

#(1+tan^2(theta))/tan(theta)#
One of trigonometric identities is the #1+tan^2(theta)=sec^2(theta)#
So the numerator would be replaced.

#sec^2(theta)/tan(theta)#

Another trigonometric identities are the
#sec(theta)=1/cos(theta)#

#tan(theta)=sin(theta)/cos(theta)#

so changing the numerator and denominator will result to

#1/cos^2(theta)/sin(theta)/cos(theta)#

#1/(cos(theta) * cancel(cos(theta)))*cancel(cos(theta))/sin(theta)#

#=color(red)(1/(cos(theta)sin(theta))#

If you understand double angle formulae...

#1/(costhetasintheta) = 1/ (1/2 * (2sinthetacostheta)) = 1/( 1/2 * sin2theta) #

#= 2/sin(2theta) = color(red)(2 csc2theta #