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Question

y=cot1[1+sinx+1sinx1+sinx1sinx] then dydx=

A
1/2
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B
2/3
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C
3
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D
1
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Solution

The correct option is A 1/2
y=cot1[1sinx+1sinx1+sinx1sinx]
Rationalising,
=cot1[(1sinx+1sinx)2(1+sinx)2(1sinx)2]
=cot1[(1+sinx+1sin)21+sinx1+sinx]
=cot1[(1+sinx+1sinx+21sinx)2sinx]
=cot1[2+2cossinx] =[1+cosxsinx=sinx1cosx]
=cot1(sinx1cosx)
=cot1[2sinx/2cosx/21(12sin2x2)]=cot1[2/sinx/2cosx/22sin2x/2]
ycot1[cot(x2)]=x2
y=x2
dydx=12
Option A is correct

1113843_1140230_ans_a3385d465cf64b5b850d2f1f687be351.jpeg

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