Question

Find the value of the trigonometric functions for the following:
tan θ = −2, θ lies in the II quadrant

Answer 1

We know that sec²θ – tan²θ = 1

sec²θ – (– 2)² = 1

sec²θ – 4 = 1

sec²θ = 1 + 4 = 5

sec θ = ${\displaystyle \pm \sqrt{5}}$

Since θ lies in the second quadrant sec θ is negative.

∴ sec θ = ${\displaystyle -\sqrt{5}}$

cos θ = ${\displaystyle \frac{1}{\sec \theta }=-\frac{1}{\sqrt{5}}}$

We know cos²θ + sin²θ = 1

${\displaystyle {(-\frac{1}{\sqrt{5}})}^2+{\sin }^2\theta }$ = 1

${\displaystyle \frac{1}{5}+{\sin }^2\theta }$ = 1

sin²θ = ${\displaystyle 1-\frac{1}{5}=\frac{5-1}{5}}$

sin²θ = ${\displaystyle \frac{4}{5}}$

sin θ = ${\displaystyle \pm \frac{2}{\sqrt{5}}}$

Since θ lies in the second quadrant sin θ is positivee.

∴ sin θ = ${\displaystyle \frac{2}{\sqrt{5}}}$

sin θ =  ${\displaystyle \frac{2}{\sqrt{5}}}$, cosec  = ${\displaystyle \frac{1}{\sin \theta }=\frac{\sqrt{5}}{2}}$

cos θ = ${\displaystyle -\frac{1}{\sqrt{5}}}$, sec θ = ${\displaystyle \frac{1}{\cos \theta }=-\sqrt{5}}$

tan θ =  – 2, cot θ = ${\displaystyle \frac{1}{\tan \theta }=-\frac{1}{2}}$