First slide

Compound angles

Question

If $α, β, γ$ are acute angles and $\cos θ = \sin β / \sin α, \cos ϕ = \sin γ / \sin α and \cos (θ - ϕ) = \sin βsin γ$ then the value of $\tan^{2} α - \tan^{2} β - \tan^{2} γ$ is equal to

Moderate

A

-1
B

0
C

1
D

2

Solution

From the third relation we get
$\begin{array}{l} \cos θcos ϕ + \sin θsin ϕ = \sin βsin γ \\ \Rightarrow \sin^{2} {θsin}^{2} ϕ = (\cos θcos ϕ - \sin βsin γ)^{2} \\ \Rightarrow (1 - \frac{\sin^{2} β}{\sin^{2} α}) (1 - \frac{\sin^{2} γ}{\sin^{2} α}) \\ = {(\frac{\sin βsin γ}{\sin^{2} α} - \sin βsin γ)}^{2} \end{array}$
$\begin{array}{l} \Rightarrow (\sin^{2} α - \sin^{2} β) (\sin^{2} α - \sin^{2} γ) = \sin^{2} {βsin}^{2} γ {(1 - \sin^{2} α)}^{2} \\ \Rightarrow \sin^{4} α (1 - \sin^{2} {βsin}^{2} γ) - \sin^{2} α \\ (\sin^{2} β + \sin^{2} γ - 2 \sin^{2} {βsin}^{2} γ) = 0 \\ ∴ \sin^{2} α = \frac{\sin^{2} β + \sin^{2} γ - 2 \sin^{2} {βsin}^{2} γ}{1 - \sin^{2} {βsin}^{2} γ} \\ and \cos^{2} α = \frac{1 - \sin^{2} β - \sin^{2} γ + \sin^{2} {βsin}^{2} γ}{1 - \sin^{2} {βsin}^{2} γ} \\ \Rightarrow \tan^{2} α = \frac{\sin^{2} β - \sin^{2} {βsin}^{2} γ + \sin^{2} γ - \sin^{2} {βsin}^{2} γ}{\cos^{2} β - \sin^{2} γ (1 - \sin^{2} β)} \\ = \frac{\sin^{2} {βcos}^{2} γ + \cos^{2} {βsin}^{2} γ}{\cos^{2} {βcos}^{2} γ} \\ = \tan^{2} β + \tan^{2} γ \\ \Rightarrow \tan^{2} α - \tan^{2} β - \tan^{2} γ = 0 \end{array}$

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