How do you find the exact value of #cot^2theta=csctheta+1# in the interval #0<=theta<360#?

1 Answer
Nov 17, 2016

Please see the explanation.

Explanation:

Substitute #cos^2(theta)/sin^2(theta)# for #cot^2(theta)# and #1/sin(theta)# for #csc(theta)#:

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

Multiply both sides by #sin^2(theta)#:

#cos^2(theta) = sin(theta) + sin^2(theta)#

Substitute #1 - sin^2(theta)# for #cos^2(theta)#

#1 - sin^2(theta) = sin(theta) + sin^2(theta)#

Combine like terms:

#2sin^2(theta) + sin(theta) - 1 = 0#

Factor

#(2sin(theta) - 1)(sin(theta) + 1) = 0#

Set each factor equal to zero:

#(2sin(theta) - 1) = 0 and (sin(theta) + 1) = 0#

Solve both equations for #sin(theta)#;

#sin(theta) = 1/2 and sin(theta) = -1#

The left equation happens twice; the right one only once:

#theta = sin^-1(1/2), theta = 180 - sin^-1(1/2), and theta = sin^-1(-1)#

#theta = 30^@, theta = 150^@, and theta = 270^@#