MathCastle
PrecalculusHard

Precalculus · Hard problem

Given cos⁡θ=35\cos\theta=\dfrac35 with θ\theta in QIV, use tan⁡θ2=1−cos⁡θsin⁡θ\tan\dfrac{\theta}{2}=\dfrac{1-\cos\theta}{\sin\theta} to find tan⁡θ2\tan\dfrac{\theta}{2}.

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