# Trigonometrical Ratios of (270° - θ)

What are the relations among all the trigonometrical ratios of (270° - θ)?

In trigonometrical ratios of angles (270° - θ) we will find the relation between all six trigonometrical ratios.

 We know that, sin (90° - θ) = cos θ cos (90° - θ) = sin θ tan (90° - θ) = cot θ csc (90° - θ) = sec θ sec (90° - θ) = csc θ cot (90° - θ) = tan θ and  sin (180° + θ) = - sin θ cos (180° + θ) = - cos θ tan (180° + θ) = tan θ csc (180° + θ) = -csc θ sec (180° + θ) = - sec θ cot (180° + θ) = cot θ

Using the above proved results we will prove all six trigonometrical ratios of (270° - θ).

sin (270° - θ) = sin [180° + 90° - θ]

= sin [180° + (90° - θ)]

= - sin (90° - θ), [since sin (180° + θ) = - sin θ]

Therefore, sin (270° - θ) = - cos θ, [since sin (90° - θ) = cos θ]

cos (270° - θ) = cos [180° + 90° - θ]

= cos [180° + (90° - θ)]

= - cos (90° - θ), [since cos (180° + θ) = - cos θ]

Therefore, cos (270° - θ) = - sin θ, [since cos (90° - θ) =  sin θ]

tan (270° - θ) = tan [180° + 90° - θ]

= tan [180° + (90° - θ)]

= tan (90° - θ), [since tan (180° + θ) = tan θ]

Therefore, tan (270° - θ) = cot θ, [since tan (90° - θ) = cot θ]

csc (270° - θ) = $$\frac{1}{sin (270° - \Theta)}$$

= $$\frac{1}{- cos \Theta}$$, [since sin (270° - θ) = - cos θ]

Therefore, csc (270° - θ) = - sec θ;

sec (270° - θ) = $$\frac{1}{cos (270° - \Theta)}$$

= $$\frac{1}{- sin \Theta}$$, [since cos (270° - θ) = -sin θ]

Therefore, sec (270° - θ) = - csc θ

and

cot (270° - θ) = $$\frac{1}{tan (270° - \Theta)}$$

= $$\frac{1}{cot \Theta}$$, [since tan (270° - θ) = cot θ]

Therefore, cot (270° - θ) = tan θ.

Solved examples:

1. Find the value of cot 210°.

Solution:

cot 210° = cot (270 - 60)°

= tan 60°; since we know, cot (270° - θ) = tan θ

= √3

2. Find the value of cos 240°.

Solution:

cos 240° = cos (270 - 30)°

= - sin 30°; since we know, cos (270° - θ) = - sin θ

= - 1/2

Trigonometric Functions