Question

tan (180-theta)???????​

Answer 1

Answer:

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.

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