Math, asked by pratimatiwari19882, 7 months ago

1÷sinAcosA[(sinA+cosA+1)(sinA+cosA-1)]

Answers

Answered by rishavkumar47

1

Answer:

2cosec²2A

Step-by-step explanation:

1÷sinAcosA[(sinA+cosA+1)(sinA+cosA-1)]

1÷sinAcosA[(sinA+cosA)²-1]

1÷sinAcosA[1+2sinAcosA-1]

1÷(sin2A)/2[Sin2A]

$\frac{2}{ {sin}^{2} (2a) }$

2cosec²2A

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