How do you solve and find the value of $tan ({cos}^{- 1} (\frac{6}{7}))$ $tan(cos^-1(6/7))$ ?

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How do you solve and find the value of $tan ({cos}^{- 1} (\frac{6}{7}))$ $tan(cos^-1(6/7))$ ?

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Answer 1 · 2017-02-01T05:34:48

0.60

Explanation:

Method 1: Use calculator
${cos}^{- 1} (\frac{6}{7}) - \to arccos (\frac{6}{7})$ $cos ^-1 (6/7) --> arccos (6/7)$
Calculator gives:
cos x = 6/7 --> $a r c x = 31^{\circ}$ $arc x = 31^@$
${tan 31}^{\circ} = 0.60$ $tan 31^@ = 0.60$
Method 2: Use right triangle OMA
with horizontal leg OA = 6
hypotenuse OM = 7
Vertical leg $A M = \sqrt{49 - 36} = \sqrt{13}$ $AM = sqrt(49 - 36) = sqrt13$
Call t the angle < AOM:
$tan t = A \frac{M}{O} A = \frac{\sqrt{13}}{6} = 0.60$ $tan t = AM/OA = sqrt13/6 = 0.60$

Answer 2 · 2017-02-01T06:21:35

$\frac{1}{\sqrt{3}}$ $1/sqrt3$

Explanation:

Let $a = {cos}^{- 1} (\frac{6}{7}) \in Q_{1}, cos a = \frac{6}{7} and tan a > 0$ $a = cos^(-1)(6/7) in Q_1, cos a = 6/7 and tan a > 0$

So the given expression

$tan a = \frac{sin a}{cos a} = \frac{\sqrt{1 - {cos}^{2} a}}{cos a} = \frac{\sqrt{1 - \frac{36}{49}}}{\frac{6}{7}} = \frac{1}{\sqrt{3}}$ $tan a = sina/cos a=sqrt(1-cos^2a)/cos a=sqrt(1-36/49)/(6/7)=1/sqrt3$

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How do you solve and find the value of $tan ({cos}^{- 1} (\frac{6}{7}))$ $tan(cos^-1(6/7))$ ?

2 Answers

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Explanation:

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