Answer:

In TRIGONOMETRICAL ratios of angles (180° - 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 (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

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