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Find the value of $x$ if $$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$$ First i tried to calculate the value of

$$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{3}{\sqrt{10}}$$ Letting

$$\theta=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{3}{\sqrt{10}}$$ taking $\sin$ both sides and using $sin(A+B)=sinAcosB+cosAsinB$ we get

$$\sin\theta=\sin\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)\cos\left(\sin^{-1}\frac{3}{\sqrt{10}}\right)+\cos\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)\sin\left(\sin^{-1}\frac{3}{\sqrt{10}}\right)$$ so

$$\sin\theta=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}$$

Now my doubt is will $\theta=\frac{\pi}{4}$ or $\theta=\frac{3\pi}{4}$ ?

My book has taken $\theta=\frac{3\pi}{4}$

Blue
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Umesh shankar
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3 Answers3

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The function $\sin^{-1}:[-1,1]\to[-\frac{\pi}2,\frac{\pi}2]$ is increasing. So $$\frac{\sqrt3}{2}<\frac2{\sqrt5}<1\quad\text{and}\quad\frac{\sqrt3}{2}<\frac3{\sqrt{10}}<1\qquad\implies \qquad\frac{2\pi}{3}<\theta<\pi$$

Ángel Mario Gallegos
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Use the definition of Principal Value,

$$\sec^{-1}\sqrt5=\tan^{-1}\sqrt{5-1}=\tan^{-1}2$$

If $\csc^{-1}\dfrac{\sqrt{10}}3=y,\csc y=\dfrac{\sqrt{10}}3$ and $0<y<\dfrac\pi2$

$\cot y=+\sqrt{\left(\dfrac{\sqrt{10}}3\right)^2-1}=\dfrac13\iff\tan y=3$

$\implies\csc^{-1}\dfrac{\sqrt{10}}3=\tan^{-1}3$

Like my answer in showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$,

$\tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\dfrac{2+3}{1-2\cdot3}=\pi+\left(-\dfrac\pi4\right)$

Can you take it from here?

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lab bhattacharjee
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  • you mean to say its better to convert everything to $\tan$ right? – Umesh shankar Apr 15 '16 at 08:36
  • @Umeshshankar, Yes, then you need to deal with one Trigonometric ratio. Else you need to use http://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y – lab bhattacharjee Apr 15 '16 at 08:38
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See $\sec^{-1}(\sqrt{5})+\csc^{-1}(\frac{\sqrt{10}}{3})=135$ so $x=45$ if we take $\theta=\pi/4$ then sum would be $\pi/2$ and not $\pi$ thus book has taken $\theta=3\pi/4$

Archis Welankar
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