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Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator.

I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$

So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$.

The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you.

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    $\begingroup$ So you did all that but you can't find $\tan\left( \dfrac{\pi}{4}\right)$? $\endgroup$ – Git Gud Jan 1 '14 at 2:01
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    $\begingroup$ Draw an isoceles right-angled triangle. Two of its angles are $\pi/4$. Then $\tan(\pi/4)$ is opposite divided by adjacent. These are equal, so $\tan(\pi/4)=1$. $\endgroup$ – André Nicolas Jan 1 '14 at 2:07
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    $\begingroup$ You don't have to actually compute $\sin^{-1}(\sqrt{2}/2)$ to solve this problem. Draw a right triangle with opposite side length $\sqrt{2}$ and hypotenuse length $2$. Now use the pythagorean theorem to find the length of the adjacent side; call this length $a$. Then compute $\sqrt{2}/a$. $\endgroup$ – Alex Wertheim Jan 1 '14 at 2:12
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    $\begingroup$ it's amazing that questions like these get more response that the questions which actually deserve attention $\endgroup$ – ibuprofen Jul 10 '19 at 10:52
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Hint: $$\tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)}$$

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    $\begingroup$ You mispelled 'answer'. $\endgroup$ – Git Gud Jan 1 '14 at 2:05
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    $\begingroup$ No, because he didn't also misspell $1$. $\endgroup$ – John Jan 1 '14 at 2:14
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$\hskip2in$ enter image description here

Using the triangle above...& the fact that $$\tan x = \frac{\text{opp}}{\text{adj}}, \space \tan \left(\frac{\pi}{4}\right)=...$$

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$tan(sin^{-1}\frac{1}{\sqrt{2}})$. So $sin(45)=\frac{1}{\sqrt{2}},x=45$ so $tan(45)=1$ hence done . Now $sin(45)=cos(45)$ thus $tan(45)=1$

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You have

$ \sin x=\frac{\sqrt{2}}{2} $

so

$ \cos x=\sqrt{1-(\sin x)^{2}}=\frac{\sqrt{2}}{2} $

and

$ \tan x =\frac{\sin x}{\cos x}\ = 1 $

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