prove the All sin, cos, Tan, cosec, sec, cot teta degrees up to 180​

1 thought on “prove the All sin, cos, Tan, cosec, sec, cot teta degrees up to 180​”

1. [tex]\mathfrak\blue{Answer \ :-}[/tex]

   • 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

   _________________________

   [tex]\large{\boxed{\boxed{\rm{Proving \ them : – }}}}[/tex]

   • 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)

   = 1−sinΘ, [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)

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

   Therefore, cot (180° + θ) = cot θ

   Reply