# 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 (180° - θ).

sin (270° + θ) = sin [1800 + 90° + θ]

= sin [1800 + (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 [1800 + 90° + θ]

= cos [I 800 + (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 [1800 + 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 csc 315°.

Solution:

csc 315° = sec (270 + 45)°

= - sec 45°; since we know, csc (270° + θ) = - sec θ

= - √2

2. Find the value of cos 330°.

Solution:

cos 330° = cos (270 + 60)°

= sin 60°; since we know, cos (270° + θ) = sin θ

= $$\frac{√3}{2}$$

Trigonometric Functions

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