Love for trigonometry-3

Geometry Level 2

Let, sin ⁡ θ + tan ⁡ θ = α \sin\theta+\tan\theta=\alpha and tan ⁡ θ − sin ⁡ θ = β \tan\theta-\sin\theta=\beta

Then ( α 2 − β 2 ) 2 α β = \dfrac{(\alpha^2-\beta^2)^2}{\alpha\beta}= ?


The answer is 16.

Fahim Muhtamim by Fahim Muhtamim , 17, Bangladesh

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2 solutions

( α 2 − β 2 ) 2 α β = ( α + β ) 2 ( α − β ) 2 tan ⁡ 2 θ − sin ⁡ 2 θ = ( 4 tan ⁡ 2 θ ) ( 4 sin ⁡ 2 θ ) sin ⁡ 2 θ ( sec ⁡ 2 θ − 1 ) = 16 tan ⁡ 2 θ sin ⁡ 2 θ sin ⁡ 2 θ tan ⁡ 2 θ = 16 \quad\dfrac{(\alpha^2-\beta^2)^2}{\alpha\beta}\\=\dfrac{(\alpha+\beta)^2(\alpha-\beta)^2}{\tan^2\theta-\sin^2\theta}\\=\dfrac{(4\tan^2\theta)(4\sin^2\theta)}{\sin^2\theta(\sec^2\theta-1)}\\=\dfrac{16\tan^2\theta\sin^2\theta}{\sin^2\theta\tan^2\theta}\\=\boxed{16}

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α = sin ⁡ θ + tan ⁡ θ \alpha=\sin \theta + \tan \theta ⇒ \Rightarrow α 2 = sin ⁡ 2 θ + 2 sin ⁡ θ tan ⁡ θ + tan ⁡ 2 θ \alpha^2=\sin^2 \theta + 2\sin \theta \tan \theta + \tan^2 \theta

β = tan ⁡ θ − sin ⁡ θ \beta=\tan \theta - \sin \theta ⇒ \Rightarrow β 2 = tan ⁡ 2 θ − 2 sin ⁡ θ tan ⁡ θ + sin ⁡ 2 θ \beta^2=\tan^2 \theta - 2\sin \theta \tan \theta + \sin^2 \theta

α 2 − β 2 = 4 sin ⁡ θ tan ⁡ θ \alpha^2-\beta^2=4\sin \theta \tan \theta

( α 2 − β 2 ) 2 = 16 sin ⁡ 2 θ tan ⁡ 2 θ (\alpha^2-\beta^2)^2=16\sin^2 \theta \tan^2 \theta

α β = tan ⁡ 2 θ − sin ⁡ 2 θ \alpha \beta = \tan^2 \theta - \sin^2 \theta

However, tan ⁡ 2 θ = sec ⁡ 2 θ sin ⁡ 2 θ \tan^2 \theta = \sec^2 \theta \sin^2 \theta

So,

α β = sec ⁡ 2 θ sin ⁡ 2 θ − sin ⁡ 2 θ = sin ⁡ 2 θ ( sec ⁡ 2 θ − 1 ) \alpha \beta = \sec^2 \theta \sin^2 \theta - \sin^2 \theta=\sin^2 \theta(\sec^2 \theta-1)

However, tan ⁡ 2 θ = s e c 2 θ − 1 \tan^2 \theta=sec^2 \theta -1

So,

α β = sin ⁡ 2 θ tan ⁡ 2 θ \alpha \beta =\sin^2 \theta \tan^2 \theta

Finally,

( α 2 − β 2 ) 2 α β = 16 sin ⁡ 2 θ tan ⁡ 2 θ sin ⁡ 2 θ tan ⁡ 2 θ = 16 \dfrac{(\alpha^2-\beta^2)^2}{\alpha \beta}=\dfrac{16\sin^2 \theta \tan^2 \theta}{\sin^2 \theta \tan^2 \theta}=\color{#69047E}\large{\boxed{16}}

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