(a)[3]
Show that $\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = 2\tan\theta$.
(b)[3]
Hence solve the equation $\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = 5\tan^3\theta$ for $-90^\circ < \theta < 90^\circ$.
Mathematics 9709 · AS & A Level · Trigonometry
Trigonometry — practice question
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