Question

simplify
sin(0) tan(180°+0)tan (90°+0)​

Answer 1

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)

= 1sinΘ, [since sin (180° - θ) = sin θ]

Therefore, csc (180° - θ) = csc θ;

sec (180° - θ) = 1cos(180°−Θ)

= 1−cosΘ, [since cos (180° - θ) = - cos θ]

Therefore, sec (180° - θ) = - sec θ

and

cot (180° - θ) = 1tan(180°−Θ)

= 1−tanΘ, [since tan (180° - θ) = - tan θ]

Therefore, cot (180° - θ) = - cot θ.