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
11 and 12 Grade Math
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