# value of $2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$

If $$x^3+bx^2+cx+1=0$$ has only real root $$\alpha$$.

Where $$(b. Then $$\displaystyle 2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$$ is

Plan

$$\tan^{-1}\bigg(\frac{2\csc \alpha}{1-\csc^2\alpha}\bigg)+\tan^{-1}\bigg(2\sin \alpha\sec^2\alpha\bigg)$$

$$\tan^{-1}\bigg(\frac{\frac{2\csc\alpha}{1-\csc^2\alpha}+2\sin\alpha\sec^2\alpha}{1-\frac{2\csc\alpha}{1-\csc^2\alpha}2\sin\alpha\sec^2\alpha}\bigg)$$

How do i solve it Help me please

Let $$f(x)=x^3+bx^2+cx+1$$

$$f(0)=1>0$$

$$f(-1)=(b-c)<0$$

So, $$\alpha$$ lies between $$f(0)$$ and $$f(-1)$$ which is $$-1,0$$

$$2\tan^{-1}(\csc\alpha)+\tan^{-1}(2\sin\alpha\sec^2\alpha)$$

$$2\tan^{-1}\alpha\left(\dfrac{1}{\sin\alpha}\right)+\tan^{-1}\left(\dfrac{2\sin\alpha}{\cos^2\alpha}\right)$$

$$2\tan^{-1}\alpha\left(\dfrac{1}{\sin\alpha}\right)+\tan^{-1}\left(\dfrac{2\sin\alpha}{1-\sin^2\alpha}\right)$$

$$2\left[\tan^{-1}\left(\dfrac{1}{\sin\alpha}\right)\right]+\tan^{-1}(\sin\alpha)$$

$$2\left(-\dfrac{\pi}{2}\right)=-\pi$$

Note that $$2\sin\alpha\sec^2\alpha=\frac{2\sin\alpha}{\cos^2\alpha}=\frac{2\sin\alpha}{1-\sin^2\alpha}$$ If $$f(x)=\arctan\frac{2x}{1-x^2}$$ then $$f'(x)=\frac{2}{1+x^2}$$ so $$f(x)$$ differs from $$2\arctan x$$ by a constant over $$(-1,1)$$. Since $$f(0)=0=2\arctan0$$, we can say that $$\arctan\frac{2\sin\alpha}{1-\sin^2\alpha}=2\arctan\sin\alpha$$ For $$x>0$$, $$\arctan(1/x)=\pi/2-\arctan x$$, so your expression evaluates to $$2\left(\frac{\pi}{2}-\arctan\sin\alpha\right)+2\arctan\sin\alpha=\pi$$ if $$\sin\alpha>0$$.

If $$\sin\alpha<0$$, the expression evaluates to $$-\pi$$, because for $$x<0$$ one has $$\arctan(1/x)=-\pi/2-\arctan x$$.

$$2\arctan p=\begin{cases} \arctan\dfrac{2p}{1-p^2} &\mbox{if } p^2<1 \\ \pi+ \arctan\dfrac{2p}{1-p^2} & \mbox{if } p>1\\-\pi+ \arctan\dfrac{2p}{1-p^2} & \mbox{if } p<-1\end{cases}$$

So, if $$2m\pi\alpha>0>(2m-1)\pi,\csc\alpha<0\implies\csc\alpha<-1$$

Consequently, $$2\arctan(\csc\alpha)=-\pi+\arctan\dfrac{2\csc\alpha}{1-\csc^2\alpha}$$ $$=-\pi+\arctan\left(-\dfrac{2\csc\alpha}{\cot^2\alpha}\right)$$

$$=-\pi-\arctan\left(\dfrac{2\csc\alpha}{\cot^2\alpha}\right)$$

$$=-\pi-\arctan\left(\dfrac{2\sin\alpha}{\cos^2\alpha}\right)$$

Here $$-1<\alpha<0,\implies\csc\alpha<0$$