Question

sin(90°-Ø)+ sinØ/sinø-sin(90°-ø)×cos(90°-ø)-cosø/cos (90°-Ø)+cosø

Answer 1

Answer:

1

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

As 90°-Ø lies in the first quadrant, all are positive and convert into co-functions i.e. sine ⇒ cosine, tangent ⇒ cotangent, etc.

sin(90°-Ø) = cos(Ø)

cos(90°-Ø) = sin(Ø)

sin(90°-Ø) + (sinØ/sinø) - [sin(90°-ø) × cos(90°-ø)] - [cosø/cos (90°-Ø)] + cosø

(BODMAS)

⇒ cos(Ø) + 1 + [cos(Ø)sin(Ø)] - [cos(Ø)/sin(Ø)] + cos(Ø)

⇒ 2cos(Ø) + 1  + cos(Ø)sin(Ø) - cot(Ø)

Hoping that I have not made any mistakes, You're welcome.

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GeniusH