Question

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

Answer 1

$\mathfrak\blue{Answer \ :-}$

• 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

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$\large{\boxed{\boxed{\rm{Proving \ them : - }}}}$

• 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