If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+1=0 (\alpha>\beta)$, then find following

Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+1=0 (\alpha>\beta)$, then find the value of $f(\alpha,\beta)=\dfrac{\beta^3}{2}\mathrm{cosec}^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)+\dfrac{\alpha^3}{2}\sec^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\alpha}{\beta}\right)$

$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\cot^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)\right)+\dfrac{\alpha^3}{2}\left(1+\tan^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\alpha}{\beta}\right)\right)$$

As $\dfrac{\alpha}{\beta}$ is positive

$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\mathrm{cosec}^2\left(1+\cot^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)\right)+\dfrac{\alpha^3}{2}\left(1+\tan^2\left(\dfrac{1}{2}\cot^{-1}\dfrac{\beta}{\alpha}\right)\right)$$

$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\cot^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)\right)+\dfrac{\alpha^3}{2}\left(1+\tan^2\left(\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)\right)$$

Let $\tan^{-1}\dfrac{\beta}{\alpha}=\theta$

$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\cot^2\dfrac{\theta}{2}\right)+\dfrac{\alpha^3}{2}\left(1+\tan^2\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\right)$$

As we have standard identity: $\tan^2\theta=\dfrac{1-\cos2\theta}{1+\cos2\theta}$

$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\dfrac{1+\cos\theta}{1-\cos\theta}\right)+\dfrac{\alpha^3}{2}\left(1+\dfrac{1-\cos\left(\dfrac{\pi}{2}-\theta\right)}{1+\cos\left(\dfrac{\pi}{2}-\theta\right)}\right)$$

$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\dfrac{1+\cos\theta}{1-\cos\theta}\right)+\dfrac{\alpha^3}{2}\left(1+\dfrac{1-\sin\theta}{1+\sin\theta}\right)$$

$$f(\alpha,\beta)=\dfrac{\beta^3}{1-\cos\theta}+\dfrac{\alpha^3}{1+\sin\theta}$$

As $\dfrac{\beta}{\alpha}=\tan\theta$

$$\cos\theta=\dfrac{\alpha}{\sqrt{\alpha^2+\beta^2}},\sin\theta=\dfrac{\beta}{\sqrt{\alpha^2+\beta^2}}$$

$$f(\alpha,\beta)=\dfrac{\beta^3}{1-\dfrac{\alpha}{\sqrt{\alpha^2+\beta^2}}}+\dfrac{\alpha^3}{1+\dfrac{\beta}{\sqrt{\alpha^2+\beta^2}}}$$

$$f(\alpha,\beta)=\dfrac{\beta^3\cdot\sqrt{\alpha^2+\beta^2}}{{\sqrt{\alpha^2+\beta^2}}-\alpha}+\dfrac{\alpha^3\sqrt{\alpha^2+\beta^2}}{\sqrt{\alpha^2+\beta^2}+\beta}$$

But actual answer is $$(\alpha^2+\beta^2)(\alpha+\beta)$$

Answer 1

$$f(\alpha,\beta)=\dfrac{\beta^3\sqrt{\alpha^2+\beta^2}\left(\sqrt{\alpha^2+\beta^2}+\alpha\right)}{\alpha^2}+\dfrac{\alpha^3\sqrt{\alpha^2+\beta^2}\left(\sqrt{\alpha^2+\beta^2}-\beta\right)}{\alpha^2}$$

$$f(\alpha,\beta)=\beta\left(\alpha^2+\beta^2+\alpha\sqrt{\alpha^2+\beta^2}\right)+\alpha\left(\alpha^2+\beta^2-\beta\sqrt{\alpha^2+\beta^2}\right)$$

$$f(\alpha,\beta)=\beta\left(\alpha^2+\beta^2\right)+\alpha\left(\alpha^2+\beta^2\right)$$ $$f(\alpha,\beta)=(\alpha+\beta)(\alpha^2+\beta^2)$$

So previously I forgot to rationalize, hence closing this.