Can someone help me solve this? Thanks!

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2 Answers
Sep 25, 2017

To verify an identity, one should make changes to only 1 side, until it is identical to the other side; I shall make changes to only the left.

Explanation:

Given:

#sec(theta)/(csc(theta)-cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#

Multiply the first fraction by 1 in the form #(csc(theta)+cot(theta))/(csc(theta)+cot(theta))#

#sec(theta)/(csc(theta)-cot(theta))(csc(theta)+cot(theta))/(csc(theta)+cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#

This makes the denominator become the difference of two squares:

#(sec(theta)(csc(theta)+cot(theta)))/(csc^2(theta)-cot^2(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#

This particular difference of two squares is the left side of the identity #csc^2(theta)-cot^2(theta) = 1#, therefore, the denominator becomes 1 and disappears:

#sec(theta)(csc(theta)+cot(theta))-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#

Use the identities #sec(theta)csc(theta) =1# and #sec(theta)cot(theta) = csc(theta)#:

#1+csc(theta)-sec(theta)/(csc(theta)+cot(theta)) = 2csc(theta)#

Multiply the next fraction by 1 in the form #(csc(theta)-cot(theta))/(csc(theta)-cot(theta))#

#1+csc(theta)-sec(theta)/(csc(theta)+cot(theta))(csc(theta)-cot(theta))/(csc(theta)-cot(theta)) = 2csc(theta)#

The produces the same difference of two squares, therefore, we shall merely delete the numerators:

#1+csc(theta)-sec(theta)(csc(theta)-cot(theta)) = 2csc(theta)#

Use the identities #sec(theta)csc(theta) =1# and #sec(theta)cot(theta) = csc(theta)#:

#1+csc(theta)-1+csc(theta) = 2csc(theta)#

Combine like terms:

#2csc(theta) = 2csc(theta)# Q.E.D.

Sep 27, 2017

Please see below

Explanation:

#sectheta/(csctheta-cottheta)-sectheta/(csctheta+cottheta)#

= #(sectheta(csctheta+cottheta)-sectheta(csctheta-cottheta))/(csc^2theta-cot^2theta)#

= #sectheta(csctheta+cottheta-csctheta+cottheta)# #-># as #csc^2theta-cot^2theta=1#

= #secthetaxx2cottheta#

= #1/costhetaxx2costheta/sintheta#

= #2/sintheta=2csctheta#