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The question is:

Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$

I did the question without using the Hint, but I don't know how to do it using the hint.

Quick working out of what I've done:

\begin{aligned} \text { If } \theta &=\tan ^{-1} 2 \\ \tan \theta &=2 \\ 0 & < \theta < \frac{\pi}{2} \end{aligned}

\begin{aligned} \cos 2 \theta &=2 \cos ^{2} \theta-1 \\ &=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\ &=\frac{3}{5} \\ 2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi \end{aligned}

\begin{array}{l} 2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\ \text { Note: } \cos ^{-1} x \text { has point symmetry } \\ \text { in }\left(0, \frac{\pi}{2}\right) \text { . } \end{array}

$$ \begin{array}{l} \cos ^{-1} x+\cos ^{-1}(-x)=\pi \\ \cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\ \therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \end{array} $$

But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !

Freddie
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5 Answers5

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$$\tan \alpha =2 \implies \tan 2\alpha = \frac{2\cdot 2}{1-2^2}=-\frac{3}{4} \implies tan(\pi-2\alpha)=\frac{3}{4} \\\implies \cos (\pi-2\alpha)=\frac 35. \blacksquare$$

Neat Math
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Note \begin{align} & \tan(2\tan^{-1}2)-\tan(\pi-\cos^{-1}\frac35)\\ =&\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}+\tan(\cos^{-1}\frac35) =\frac{2\cdot2}{1-2^2}+ \frac43=0 \end{align} which leads to $$2\tan^{-1}2=\pi-\cos^{-1}\frac35$$

Quanto
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It isn't necessary to use the hint. What you have done is correct and valid! Nonetheless, here's the intended method using hint.

Let $\theta=2\tan^{-1}2$ and $\alpha=\pi-\cos^{-1}\frac 3 5$. Then, $$\tan\theta=\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}$$ $$\tan\theta=-\frac{4}{3}\quad (*)$$ and using the hint, $$\tan\alpha=-\tan(\cos^{-1}\frac{3}{5})$$ Using $\cos^{-1}\frac{A}{\sqrt{A^2+B^2}}=\tan^{-1}\frac{B}{A}$, we get $$\tan\alpha=-\tan(\tan^{-1}\frac{4}{3})$$ $$\tan\alpha=-\frac 4 3\quad (**)$$ From $(*)$ and $(**)$, we conclude $$\theta=\alpha$$ Or $\text{LHS}=\text{RHS}$ as desired.

Hope this is clear :)

ultralegend5385
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Let $\cos^{-1}\dfrac35=y,\cos y=?$

Using Principal values, as $1>\dfrac35>0; 0<y<\dfrac\pi2$

$\tan y=\dfrac{+\sqrt{1-\left(\dfrac35\right)^2}}{\dfrac35}=?, y=\tan^{-1}\dfrac43$

$$\pi-2\tan^{-1}2=2\left(\dfrac\pi2-\tan^{-1}2\right)=2\cot^{-1}2$$

Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

and

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 $xy>1$

lab bhattacharjee
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Hint:

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 $xy>1$

$$2\tan^{-1}2=\pi+\tan^{-1}\dfrac{2\cdot2}{1-2^2}=\pi+\tan^{-1}\left(?\right)$$

Now using Principal values $\tan^{-1}(-y)=-\tan^{-1}y$

Again, if $\tan^{-1}\dfrac43=u,0<u<\dfrac\pi2$ as $\dfrac43>0$

$\cos u=\dfrac1{+\sqrt{1+\tan^2y}}=?$

Can you take it home from here?

lab bhattacharjee
  • 274,582