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

Question

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 $(x^2-y^2)^2=16 xy$.

Answer 1

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 $(x^2 - y^2)^2 = 16xy$.

Answer 2

Hint:

Let us start without knowing the proposition to be established.

Solve for $\tan\theta,\sin\theta$

Now use $$\tan^2\theta=\dfrac{\sin^2\theta}{1-\sin^2\theta}$$

and simplify.