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Lemma
Trigonometric Identities Difficulty 4

TMUA practice: Trig Identities, problem 5

A TMUA-calibre trig identities problem at difficulty 4 of 5, with the full worked solution.

The question
Let θ\theta be an angle in the second quadrant (90∘≤θ<180∘90^\circ \le \theta < 180^\circ) with tan⁡θ=−23\tan\theta = -\tfrac23. Then the value of tan⁡(90∘+θ)+cos⁡(180∘+θ)sin⁡(270∘−θ)−cot⁡(−θ)\dfrac{\tan(90^\circ + \theta) + \cos(180^\circ + \theta)}{\sin(270^\circ - \theta) - \cot(-\theta)} is
A 2+132−13\dfrac{2 + \sqrt{13}}{2 - \sqrt{13}}
B 2−132+13\dfrac{2 - \sqrt{13}}{2 + \sqrt{13}}
C 2+392−39\dfrac{2 + \sqrt{39}}{2 - \sqrt{39}}
D 2+3132 + 3\sqrt{13}

The correct answer is highlighted. A

Worked solution

Reduce the allied angles. Using the standard relations,

tan⁡(90∘+θ)=−cot⁡θ,cos⁡(180∘+θ)=−cos⁡θ,sin⁡(270∘−θ)=−cos⁡θ,cot⁡(−θ)=−cot⁡θ.\tan(90^\circ + \theta) = -\cot\theta, \quad \cos(180^\circ + \theta) = -\cos\theta, \quad \sin(270^\circ - \theta) = -\cos\theta, \quad \cot(-\theta) = -\cot\theta.

So the numerator is −cot⁡θ−cos⁡θ-\cot\theta - \cos\theta, and the denominator is −cos⁡θ−(−cot⁡θ)=cot⁡θ−cos⁡θ-\cos\theta - (-\cot\theta) = \cot\theta - \cos\theta. The expression becomes

−cot⁡θ−cos⁡θcot⁡θ−cos⁡θ=− cot⁡θ+cos⁡θcot⁡θ−cos⁡θ.\frac{-\cot\theta - \cos\theta}{\cot\theta - \cos\theta} = -\,\frac{\cot\theta + \cos\theta}{\cot\theta - \cos\theta}.

Values of the functions. In the second quadrant sin⁡θ>0\sin\theta > 0 and cos⁡θ<0\cos\theta < 0. From tan⁡θ=−23\tan\theta = -\tfrac23 (a 22-33-13\sqrt{13} triangle),

sin⁡θ=213,cos⁡θ=−313,cot⁡θ=cos⁡θsin⁡θ=−32.\sin\theta = \frac{2}{\sqrt{13}}, \qquad \cos\theta = -\frac{3}{\sqrt{13}}, \qquad \cot\theta = \frac{\cos\theta}{\sin\theta} = -\frac32.

Substitute.

− cot⁡θ+cos⁡θcot⁡θ−cos⁡θ=− −32−313−32+313=− 12+11312−113,-\,\frac{\cot\theta + \cos\theta}{\cot\theta - \cos\theta} = -\,\frac{-\tfrac32 - \tfrac{3}{\sqrt{13}}}{-\tfrac32 + \tfrac{3}{\sqrt{13}}} = -\,\frac{\tfrac12 + \tfrac{1}{\sqrt{13}}}{\tfrac12 - \tfrac{1}{\sqrt{13}}},

after cancelling the common factor −3-3. Multiplying numerator and denominator by 2132\sqrt{13}:

− 13+213−2=13+22−13=2+132−13.-\,\frac{\sqrt{13} + 2}{\sqrt{13} - 2} = \frac{\sqrt{13} + 2}{2 - \sqrt{13}} = \frac{2 + \sqrt{13}}{2 - \sqrt{13}}.

Answer: A.