Question

prove ;

sin(90°-Ã) = cosÃ​

Answer 1

sin (90° + θ) = cos θ

cos (90° + θ) = OFOEOFOE

cos (90° + θ) = −DCOC−DCOC, [OF = -DC and OE = OC, since ∆ OCD ≅ ∆ OEF]

cos (90° + θ) = - sin θ.

tan (90° + θ) = FEOFFEOF

tan (90° + θ) = OD−DCOD−DC, [FE = OD and OF = - DC, since ∆ OCD ≅ ∆ OEF]

tan (90° + θ) = - cot θ.

Similarly, csc (90° + θ) = 1sin(90°+Θ)1sin(90°+Θ)

csc (90° + θ) = 1cosΘ1cosΘ

csc (90° + θ) = sec θ.

sec (90° + θ) = 1cos(90°+Θ)1cos(90°+Θ) 

sec (90° + θ) = 1−sinΘ1−sinΘ

sec (90° + θ) = - csc θ.

and cot (90° + θ) = 1tan(90°+Θ)1tan(90°+Θ)

cot (90° + θ) = 1−cotΘ1−cotΘ

cot (90° + θ) = - tan θ.

Answer 2

see the attachment

hope this helps you

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