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Question

If (tan1x)2+(cot1x)2=5π28, then x=

A
1
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Solution

The correct option is A 1
Let tan1x=y
Therefore
y2+(π2y)2=5π28
2y22πy2+π24=5π28
2y2πy3π28=0
y=π4 and y=3π4
Hence
tan1(x)=π4
x=1 and
tan1(x)=3π4
x=1
However only x=1 satisfies the above quadratic equation.

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