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Evaluate tan(cos ^-1 3/5) and assume that all angles are in Quadrant I.

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\bf cos^(-1)\left( (3)/(5) \right)=\theta \quad \stackrel{\textit{this simply means}}{\implies }\quad cos(\theta )=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}

and with that, let us find the opposite side, keeping in mind that we're in the I Quadrant, and thus the opposite side is positive, just like adjacent as well,


\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm√(c^2-a^2)=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm√(5^2-3^2)=b\implies \pm√(16)=b\implies \pm 4=b\implies \stackrel{I~Quadrant}{+4=b}\\\\ -------------------------------\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{4}}{\stackrel{adjacent}{3}}
User Callebe
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8.5k points
2 votes

Answer:

Value of
tan\,(cos^(-1)\,(3)/(5))=(4)/(3)

Explanation:

Given:


tan\,(cos^(-1)\,(3)/(5))

All angles are in Quadrant 1.

To find: Value of given Expression.

Let,


cos^{-1\,(3)/(5)}=\theta

Now, To find
tan\,\theta


\implies cos\,\theta=(3)/(5)

we know that,


tan\,\theta=(Opposite)/(Adjacent)

Using right angled triangle and for
cos\,\theta=(3)/(5)

Figure is attached.

By Pythagoras theorem,

AC² = AB² + CB²

5² = AB² + 3²

AB² = 25 - 9

AB² = 16

AB = 4

So,
tan\,\theta=(Opposite)/(Adjacent)=(4)/(3)

Therefore, Value of
tan\,(cos^(-1)\,(3)/(5))=(4)/(3)

Evaluate tan(cos ^-1 3/5) and assume that all angles are in Quadrant I.-example-1
User Tete
by
8.4k points

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