Math, asked by nancyinsan1969, 1 year ago

(1/sec^2a-cos^a+1/cosec^2a-sin^a)cos^2asin^2a=1-cos^2asin^2a/1-cos^2asin^2a

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Answered by ajayshotia

1

(1/sec²a-cos²a+1/cosec²a-sin²a)cos²asin²a
=(cos²a-cosa+sin²a-sina)cos²asin²a
=(1-cosa-sina)cos²asin²a
=

Answered by sandy1816

0

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