If $a \cos^2 \theta + b \sin^2 \theta = c$, show that \(\tan^2 \theta = \frac{a - c}{c -

Question

If $a \cos^2 \theta + b \sin^2 \theta = c$, show that $\tan^2 \theta = \frac{a - c}{c - b}$.[4] Given that $\sqrt{ (2 \cos^2 x - \sin x) } = 2 \sin x$, show that $6 \sin^2 x + \sin x - 2 = 0$.[2] Hence, solve the equation $\sqrt{ (2 \cos^2 x - \sin x) } = 2 \sin x$ for $0\le x\le2\pi$.[4]

Accepted Answer

(a) $\tan^2\theta=\frac{a-c}{c-b}$ (shown) (b)(i) $6\sin^2x+\sin x-2=0$ (shown) (b)(ii) $x=\frac\pi6$ or $\frac{5\pi}{6}$

2024 CCHY P1 Q11

2024 CCHY P1 Q11

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