Question

Suppose 0 is an angle in the standard position whose terminal side is in Quadrant IV and cot0= -6/7 Find the exact values of the five remaining trigonometric functions of 0

Answer 1

$\sin \theta=\dfrac{p}{h} =\dfrac{-7}{\sqrt{85} }$, $\cos \theta=\dfrac{b}{h} =\dfrac{6}{\sqrt{85}}$, $\tan \theta=\dfrac{p}{b} =\dfrac{-7}{6}$,

$\sec \theta=\dfrac{h}{b} =\dfrac{\sqrt{85} }{6}$ and $\csc \theta=\dfrac{h}{p} =\dfrac{-\sqrt{85}}{ 7}$

Explanation:

Given,

$\cot \theta= \dfrac{-6}{7}$

To find, the exact values of the five remaining trigonometric functions of $\theta$ = ?

We know that,

$\cot \theta= \dfrac{-6}{7}=\dfrac{b}{p}$

Where, b = base and p = perpendicular

By Pythagoras's theorem,

Hypotaneous, $h=\sqrt{p^2+b^2}$

$=\sqrt{(-6)^2+7^2}=\sqrt{36+49} =\sqrt{85}$

In IVth quadrant,

$\cot \theta$ and $\sec \theta$ are positive.

$\sin \theta=\dfrac{p}{h} =\dfrac{-7}{\sqrt{85} }$, $\cos \theta=\dfrac{b}{h} =\dfrac{6}{\sqrt{85}}$, $\tan \theta=\dfrac{p}{b} =\dfrac{-7}{6}$,

$\sec \theta=\dfrac{h}{b} =\dfrac{\sqrt{85} }{6}$ and $\csc \theta=\dfrac{h}{p} =\dfrac{-\sqrt{85}}{ 7}$