# Trigonometric ratios of minus theta(−Θ)

In this section we will discuss the relation among all trigonometric ratios of minus theta (-Θ). Here we will find the relation between all trigonometrical ratios.
Let a ray OX revolve anti clockwise direction to make an angle $\angle XOA = \Theta$.
Let P(x,y) be a point on ray OA such that OP=r. Draw PM perpendicular from P on the X-axis . Let there be another rotation of ray OX in clockwise direction to make an angle $\angle XOA' = -\Theta$. Let P' be a point on OA' such that OP = OP'. Draw P'M perpendicular from P' on X-axis.
Consider two triangles $\Delta$ POM and $\Delta$ P'OM.

$angle$ POM = $angle$ P'OM ----(same magnitude)

OM = OM ----(reflexive)
$angle$ PMO = $angle$P'M ----(each 90 degrees)

∴ $\Delta POM \cong \Delta$ P'OM ----( By ASA rule)

∴ according to the trigonometric sign rule we get,

P'M = - PM and OP = OP'.

By definition of trigonometric ratios,

$sin (- \Theta) = \frac{P'M}{OP'}$

$sin (- \Theta) = -\frac{PM}{OP}$

$sin (- \Theta) = - sin ( \Theta)$

Now, $cos (- \Theta) = \frac{OM}{OP'}$

$cos (- \Theta) = \frac{OM}{OP}$ ----(since OP = OP')

$cos (- \Theta) = cos ( \Theta)$

Again, $tan (- \Theta) = \frac{P'M}{OM}$

$tan (- \Theta) = -\frac{PM}{OM}$ ----(since P'M = - PM)

$tan (- \Theta) = -tan ( \Theta)$

As we know that, $csc ( \Theta) = \frac{1}{sin \Theta}$

⇒ $csc (- \Theta) = \frac{1}{-sin \Theta}$

⇒ $csc (- \Theta) = - csc ( \Theta)$

Similarly, $sec( \Theta) = \frac{1}{cos \Theta}$

⇒ $sec (- \Theta) = \frac{1}{cos \Theta}$

⇒ $sec (- \Theta) = (sec \Theta)$

And, $cot( \Theta) = \frac{1}{tan \Theta}$

⇒ $cot (- \Theta) = \frac{1}{-tan \Theta}$

⇒ $cot (- \Theta) = -(cot \Theta)$

## Examples on Trigonometric ratios of minus theta(−Θ)

1) Find the value of sin (-$30)^{0}$
Solution : As we know that,
$sin (- \Theta) = - sin ( \Theta)$

Here $\Theta = 30^{0}$ and $sin (30)^{0} = \frac{1}{2}$

$sin (- 30^{0}) = - sin 30^{0}$

$sin (- 30^{0}) = -\frac{1}{2}$

2) Find the value of cos (-$30)^{0}$
Solution : As we know that,
$cos (- \Theta) = cos ( \Theta)$

Here $\Theta = 30^{0}$ and $cos (30)^{0} = \frac{\sqrt{3}}{2}$

$cos (- 30^{0}) = cos 30^{0}$

$cos (- 30^{0}) = \frac{\sqrt{3}}{2}$

3) Find the value of tan (-$30)^{0}$
Solution : As we know that,
$tan (- \Theta) = - tan ( \Theta)$

Here $\Theta = 30^{0}$ and $tan (30)^{0} = \frac{1}{\sqrt{3}}$

$tan (- 30^{0}) = - tan 30^{0}$

$tan (- 30^{0}) = -\frac{1}{\sqrt{3}}$