Question

If acos^2(θ) + bsin^2(θ) = c. Show that tan^2(θ)= (a-c)/(c-b)

Answer 1

Answer on Question #79004 - Math - Trigonometry

$$\begin{array}{l} a \cos^{2} \theta + b \sin^{2} \theta = a \left(\cos^{2} \theta + \sin^{2} \theta\right) + (b - a) \sin^{2} \theta = a + (b - a) \sin^{2} \theta, \\ \Rightarrow a + (b - a) \sin^{2} \theta = c \end{array}$$

i.e.

$$\sin^{2} \theta = \frac{c - a}{b - a}$$ $$\begin{array}{l} a \cos^{2} \theta + b \sin^{2} \theta = (a - b) \cos^{2} \theta + b \left(\cos^{2} \theta + \sin^{2} \theta\right) = b + (a - b) \cos^{2} \theta, \\ \Rightarrow b + (a - b) \cos^{2} \theta = c \end{array}$$

i.e.

$$\cos^{2} \theta = \frac{c - b}{a - b}$$

Therefore

$$\tan^{2} \theta = \frac{\sin^{2} \theta}{\cos^{2} \theta} = \frac{a - c}{c - b}$$

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