sin(90-A)cosA+sinA.cos(90-A) find the value
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Answer:
Sina.cosa - {sina.cos(90°-a).cosa}/{sec(90°-a)} - {cosa.sin(90°-a)sina}/{cosec(90°-a)}
= sina.cosa - {sina.sina.cosa}/{coseca} - {cosa.cosa.sina}/{seca} [∵ sin(90°-a)=cosa, cos(90°-a)=sina , sec(90°-a) = coseca , cosec(90°-a) = seca]
= sina.cosa - sin³a.cosa - cos³a.sina [∵1/seca = cosa , 1/coseca = sina]
= sina.cosa - sina.cosa[sin²a + cos²a]
= sina.cosa - sina.cosa × 1 [ ∵sin²Ф + cos²Ф = 1 ]
= 0
Answered by
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Hope you understand.........
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