How do you solve $1=cot^2theta+csctheta$ for $0<=theta<=2pi$ ?

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Answer 1 · 2016-09-20T13:38:40

Apply the identities $cottheta = costheta/sintheta$ and $csctheta = 1/sintheta$ :

$cos^2theta/sin^2theta + 1/sintheta = 1$

$cos^2theta/sin^2theta + sintheta/sin^2theta = 1$

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

$cos^2theta + sin theta = sin^2theta$

Apply the identity $sin^2theta + cos^2theta = 1 ->cos^2theta = 1 - sin^2theta$

$1 - sin^2theta + sin theta - sin^2theta = 0$

$-2sin^2theta + sin theta + 1 = 0$

$-2sin^2theta + 2sintheta - sin theta + 1 = 0$

$-2sintheta(sin theta - 1) - 1(sin theta - 1) = 0$

$(-2sintheta - 1)(sin theta - 1) = 0$

$sintheta = -1/2 and sin theta = 1$

$theta = (7pi)/6, (11pi)/6 and pi/2$

However, since $pi/2$ renders the equation undefined, that solution is extraneous. Hence, our solution set is ${(7pi)/6, (11pi)/6}$ .

Hopefully this helps!

How do you solve 1=cot^2theta+csctheta1=cot^2theta+csctheta for 0<=theta<=2pi0<=theta<=2pi?