Question

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

Answer 1

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

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,

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

Answer 2

Answer:

Value of $tan\,(cos^{-1}\,\frac{3}{5})=\frac{4}{3}$

Step-by-step explanation:

Given:

$tan\,(cos^{-1}\,\frac{3}{5})$

All angles are in Quadrant 1.

To find: Value of given Expression.

Let,

$cos^{-1\,\frac{3}{5}}=\theta$

Now, To find $tan\,\theta$

$\implies cos\,\theta=\frac{3}{5}$

we know that,

$tan\,\theta=\frac{Opposite}{Adjacent}$

Using right angled triangle and for $cos\,\theta=\frac{3}{5}$

Figure is attached.

By Pythagoras theorem,

AC² = AB² + CB²

5² = AB² + 3²

AB² = 25 - 9

AB² = 16

AB = 4

So, $tan\,\theta=\frac{Opposite}{Adjacent}=\frac{4}{3}$

Therefore, Value of $tan\,(cos^{-1}\,\frac{3}{5})=\frac{4}{3}$