inverse trigonometric functions

**darkangel** · April 12th 2008, 11:42 AM

Help needed on these two problems -

1. Prove that

tan^(-1) x = sin^(-1)( x / sqrt(1 + x^2))

2. Find the exact value of -
$\cos\left( \tan^{-1}(\frac {5}{12}) + \cot^{-1}(\frac {5}{12})\right)<br />$

**Mathstud28** · April 12th 2008, 12:14 PM

Originally Posted by darkangel

Help needed on these two problems -

1. Prove that

tan^(-1) x = sin^(-1)( x / sqrt(1 + x^2))

2. Find the exact value of -
$\cos\left( \tan^{-1}(\frac {5}{12}) + \cot^{-1}(\frac {5}{12})\right)$

for the second one rewrite it as $cos\bigg(arctan\bigg(\frac{5}{12}\bigg)+arctan\big g(\frac{12}{5}\bigg)\bigg)$ ..then split it apart using $cos(x+y)=cos(x)\cdot{cos(y)}+sin(x)\cdot{sin(y)}$ and use the fact that $cos(arctan(x))=\frac{1}{\sqrt{1+x^2}}$ to finish

**galactus** · April 12th 2008, 12:17 PM

The second one should be easy.

$tan^{-1}(x)+cot^{-1}(x)=\frac{\pi}{2}$

**o_O** · April 12th 2008, 12:19 PM

1. Set $y = \sin^{-1} \frac{x}{\sqrt{1+x^{2}}}$ . This implies:

$\sin y = \frac{x}{\sqrt{1+x^{2}}}$

Using the identity: $\sin^{2} x + \cos^{2} x = 1 \Rightarrow \cos x = \sqrt{1 - \sin^{2} x}$

So, find an expression of $\cos y$ .

Then, find tan y using the relation: $\tan y = \frac{\sin y}{\cos y}$ and solve for y.

**darkangel** · April 12th 2008, 11:00 PM

aite Thanks for everyone's help

**Krizalid** · April 13th 2008, 05:37 AM

Originally Posted by galactus

$tan^{-1}(x)+cot^{-1}(x)=\frac{\pi}{2}$

Just a quick note here.

Observe that $\left( {\tan ^{ - 1} x} \right)' + \left( {\cot ^{ - 1} x} \right)' = 0,$ so $\tan ^{ - 1} x - \frac{\pi }<br /> {4} + \cot ^{ - 1} x - \frac{\pi }<br /> {4} = 0,$ and the result follows.

**darkangel** · April 13th 2008, 09:44 AM

woah ! thanks Krizalid !!

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