<s>[INST] If $\tan\theta+\sin\theta=x$ and $\tan\theta-\sin\theta=y$, then $(x^2-y^2)^2=16 xy$ I have tried to prove this trig identity but I seem to be heading nowhere…

If $\tan\theta+\sin\theta=x$ and $\tan\theta-\sin\theta=y$, prove that $(x^2-y^2)^2=16 xy$.

 [/INST] Let $x = \tan(\theta) + \sin(\theta)$ and $y = \tan(\theta) - \sin(\theta)$. 
R.T.P: $(x^2 - y^2)^2 = 16xy$
Proof: LHS = $(x^2-y^2)^2 = ((x-y)(x+y))^2 = (x-y)^2(x+y)^2 = (2 \sin(\theta))^2(2 \tan(\theta))^2 = 16 \sin^{2}(\theta) \tan^{2}(\theta)$.
Note that $xy = \tan^{2}(\theta) - \sin^{2}(\theta) = \frac{\sin^{2}(\theta)[1 -\cos^{2}(\theta)]}{\cos^{2}(\theta)} = \tan^{2}(\theta) \sin^{2}(\theta).$
Hence we have that $(x^2 - y^2)^2 = 16xy$.
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