Question

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

Answer 1

`\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}}`

Answer 2

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)`