calculus - Showing that $sin x;f(sin x);f^prime(cos x)+cos x;f(cos x);f^prime(sin x)=frac{2}{pisin xcos x}$ for $f(x)$ defined by a series

Let
$$f(x) = 1 + \left(\frac12\cdot x\right)^2+\left(\frac12\cdot\frac34 \cdot x^2\right)^2+\left(\frac12\cdot\frac34\cdot\frac56\cdot x^3\right)^2+\cdots$$
Prove that

$$\sin x\;f(\sin x)\;f^\prime(\cos x)+\cos x\;f(\cos x)\;f^\prime(\sin x)=\frac{2}{\pi\sin x\cos x}$$

No hints are available. Any help would be appreciated. Thanks.

I have been trying to solve this problem by finding explicit form of $f(x)$. I have noticed that the series is similar to the power series of $1/\sqrt{1-x}$, thus I have spent a lot of time connecting it to $f(x)$. Apparently, it did not work.

I also have tried to transform the claimed equality into well-known identity of trigonometric functions, such as $\sin^2x+\cos^2x =1$.

(Edited)

Thanks Parcly Taxel for editing the article. I did not know how to ask properly.

Answer

The problem is taken from this website. This is Legendre's identity for elliptic integrals in disguise.

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We have $$f(x) = \frac{2K(x)}{\pi}\tag{1}$$ where $$K(x)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-x^2\sin^2t}},E(x)=\int_{0}^{\pi/2}\sqrt{1-x^2\sin^2t}\,dt\tag {2}$$ Now Legendre's identity states that $$K(x) E(\sqrt {1-x^2})+K(\sqrt{1-x^2})E(x)-K(x)K(\sqrt{1-x^2})=\frac{\pi}{2}\tag{3}$$ We also need the formula $$E(x) =x(1-x^2) \frac{dK(x)} {dx} +(1-x^2) K(x)\tag{4}$$ Using $(1)$ and $(4)$ the identity $(3)$ can be transformed into an identity connecting $f$ and its derivative $f'$ and the finally replacing $x$ by $\sin x$ completes the job.