Math, asked by sherlyaishu27, 1 year ago

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Answered by prudvireddy
0

Step-by-step explanation:

X = a log(sec Q)

dX / dQ = a (1 /sec Q)(tan Q sec Q)

dX / dQ = a Tan Q ---------( 1 )

Y = a ( tan Q - 1 )

dY / dQ = a ( sec^2 Q - 0)

dY / dQ = a sec^2 Q ----------( 2 )

( 2 ) ÷ ( 1 )

dY / dQ ÷ dX / dQ = a (sec^2 Q) ÷ a (tan Q)

dY / dX = 1 / ( sin Q cos Q)

dY / dX = 2 / (2 sin Q cos Q)

dY / dX = 2 / sin 2Q

dY / dX = 2 cosec 2Q

Hence proved

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