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

Where $(b<c)$. 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

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jacky
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3 Answers3

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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 $\arctan(1/x)=-\pi/2-\arctan x$.

egreg
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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 $-1,0$

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

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

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

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

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

Key Flex
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Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

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

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

lab bhattacharjee
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