Prove that tan 180° = cos 90°
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Answers
Step-by-step explanation:
Using the above proved results we will prove all six trigonometrical ratios of (180° - θ).
sin (180° - θ) = sin (90° + 90° - θ)
= sin [90° + (90° - θ)]
= cos (90° - θ), [since sin (90° + θ) = cos θ]
Therefore, sin (180° - θ) = sin θ, [since cos (90° - θ) = sin θ]
cos (180° - θ) = cos (90° + 90° - θ)
= cos [90° + (90° - θ)]
= - sin (90° - θ), [since cos (90° + θ) = -sin θ]
Therefore, cos (180° - θ) = - cos θ, [since sin (90° - θ) = cos θ]
tan (180° - θ) = cos (90° + 90° - θ)
= tan [90° + (90° - θ)]
= - cot (90° - θ), [since tan (90° + θ) = -cot θ]
Therefore, tan (180° - θ) = - tan θ, [since cot (90° - θ) = tan θ]
csc (180° - θ) = 1sin(180°−Θ)1sin(180°−Θ)
= 1sinΘ1sinΘ, [since sin (180° - θ) = sin θ]
Therefore, csc (180° - θ) = csc θ;
sec (180° - θ) = 1cos(180°−Θ)1cos(180°−Θ)
= 1−cosΘ1−cosΘ, [since cos (180° - θ) = - cos θ]
Therefore, sec (180° - θ) = - sec θ
and
cot (180° - θ) = 1tan(180°−Θ)1tan(180°−Θ)
= 1−tanΘ1−tanΘ, [since tan (180° - θ) = - tan θ]
Therefore, cot (180° - θ) = - cot θ.
Solved examples:
1. Find the value of sec 150°.
Solution:
sec 150° = sec (180 - 30)°
= - sec 30°; since we know, sec (180° - θ) = - sec θ
= - 2√32√3
2. Find the value of tan 120°.
Solution:
tan 120° = tan (180 - 60)°
= - tan 60°; since we know, tan (180° - θ) = - tan θ
= - √3