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Answer 1

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Answer 2

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

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