Trigonometric ratios of 270 degrees plus theta
(270 + 
θ) 

We will prove the trigonometric ratios of 270 degrees plus theta
(270 + $\Theta$) using the trigonometric ratios of (180 + $\Theta$).
First we will revised all the trigonometric ratios of allied angles which you have already learnt in the previous section of trigonometric functions.
sin(-$\Theta) = - sin \Theta$
cos(-$\Theta) = cos \Theta$
tan(-$\Theta) = - tan \Theta$

sin( 90 - $\Theta) = cos \Theta$
cos( 90 - $\Theta) = sin \Theta$
tan( 90 -$\Theta) = cot \Theta$
csc( 90 - $\Theta) = sec \Theta$
sec( 90 - $\Theta) = csc \Theta$
cot( 90 - $\Theta) = tan \Theta$

sin( 180 - $\Theta) = sin \Theta$
cos( 180 - $\Theta) = -cos \Theta$
tan( 180 -$\Theta) = -tan \Theta$
csc( 180 - $\Theta) = csc\Theta$
sec( 180 - $\Theta) = -sec\Theta$
cot( 180- $\Theta) = -cot \Theta$

csc(-$\Theta) = - csc \Theta$
sec(-$\Theta) = sec \Theta$
cot(-$\Theta) = - cot \Theta$

sin( 90 + $\Theta) = cos \Theta$
cos( 90 + $\Theta) = -sin \Theta$
tan( 90 + $\Theta) = -cot \Theta$
csc( 90 + $\Theta) = sec \Theta$
sec( 90 + $\Theta) = -csc \Theta$
cot( 90 + $\Theta) = -tan \Theta$

sin( 180 + $\Theta) = -sin \Theta$
cos( 180 + $\Theta) = -cos \Theta$
tan( 180 +$\Theta) = tan \Theta$
csc( 180 + $\Theta) = -csc\Theta$
sec( 180 + $\Theta) = -sec\Theta$
cot( 180+ $\Theta) = cot \Theta$

Using the above results we can easily find out the trigonometric ratios of 270 degrees plus theta (270 + $\Theta$)
1) sin( $270 + \Theta ) = sin [ 180 + ( 90 + \Theta$)]
But we know that sin(180 + $\Theta)= - sin \Theta$)
∴ sin [ 180 + ( 90 + $\Theta)] = -sin (90 +\Theta$)
But sin (90 +$\Theta) = cos \Theta$)
∴ $\underline {sin( 270 + \Theta )= - cos \Theta}$

2) cos( $270 + \Theta ) = cos [ 180 + ( 90 + \Theta$)]
But we know that cos(180 + $\Theta)= - cos \Theta$)
∴ cos [ 180 + ( 90 + $\Theta)] = -cos (90 +\Theta$)
But cos (90 +$\Theta) = -sin \Theta$)
∴ $\underline {cos( 270 + \Theta )= sin \Theta}$

3) tan( $270 + \Theta ) = tan [ 180 + ( 90 + \Theta$)]
But we know that tan(180 + $\Theta)= tan \Theta$)
∴ tan[ 180 + ( 90 + $\Theta)] = tan (90 +\Theta$)
But tan (90 +$\Theta) = -cot \Theta$)
∴ $\underline {tan( 270 + \Theta )= -cot \Theta}$

4) $csc( 270 + \Theta ) = \frac{1}{sin( 270 + \Theta )}$
But $sin( 270 + \Theta )= - cos \Theta$
∴ csc( $270 + \Theta )= \frac{1}{-cos \Theta}$
∴ $\underline {csc( 270 + \Theta )= - sec \Theta}$

5) $sec( 270 + \Theta ) = \frac{1}{cos( 270 + \Theta )}$
But $cos( 270 + \Theta )= sin \Theta$
∴ sec( $270 + \Theta )= \frac{1}{sin \Theta}$
∴ $\underline {sec( 270 + \Theta )= csc \Theta}$

6) $cot( 270 + \Theta ) = \frac{1}{tan( 270 + \Theta )}$
But $tan( 270 + \Theta )= -cot \Theta$
∴ cot( $270 + \Theta )= \frac{1}{-cot \Theta}$
∴ $\underline {cot( 270 + \Theta )= - tan \Theta}$

Examples on Trigonometric ratios of 270 degrees plus theta
(270 + θ) 

1) Find the value of $sin 300^{0}$
Solution : $sin 300^{0}$
300 = 270 + 30
As we know that $sin (270 + \Theta) = -cos\Theta$
∴ $sin 300^{0}$ = sin (270 +30) = - cos 30
⇒ $sin 300^{0} = -\frac{\sqrt{3}}{2}$

2) Find the value of $tan 330^{0}$
Solution : $tan 330^{0}$
330 = 270 + 60
As we know that $tan (270 + \Theta) = -cot\Theta$
∴ $tan 330^{0}$ = tan (270 +60) = - cot 60
⇒ $tan 330^{0} = -\frac{1}{\sqrt{3}}$


11th grade math

From Trigonometric ratios of 270 degrees plus theta (270 + θ) to Home