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Question

Solve: sin{sin1(1322)+cos1(3+122)}

A
12
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B
32
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C
0
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D
12
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Solution

The correct option is C 0
sin{sin1(1322)+cos1(3+122)}
=sin{sin1(3122)+cos1(3+122)} [Since sin1(x)=sin1x]
=sin{sin1(3122)+π2sin1(3+122)}
=sin{π2sin1(3122)sin1(3+122)}
=sin{π2π2}=0

[sin1(3122)+sin1(3+122)=sin1(3122.3122+3+122.3+122)=sin1]

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