Question
If y = tan−1√(1+sinx1−sinx),π2<x<π, then dydx equals
Answer: Option A
:
A
y=tan−1
⎷(1−cos(π2+x)1+cos(π2+x))=tan−1∣∣tan(π4+x2)∣∣⋯)(i)
Now, π2<x<π
∴ π4<x2<π2
or π2<π4+x2<3π4
∴ ∣∣tan(π4+x2)∣∣=−tan(π4+x2) (∴ in second quadrant)
=tan{π−(π4+x2)}
From Eq.(i),
y=tan−1tan{π−(π4+x2)}
=π−(π4+x2)
=3π4−x2
(∵principal value of tan−1 x in−π2 to π2)
∴ dydx=−12
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:
A
y=tan−1
⎷(1−cos(π2+x)1+cos(π2+x))=tan−1∣∣tan(π4+x2)∣∣⋯)(i)
Now, π2<x<π
∴ π4<x2<π2
or π2<π4+x2<3π4
∴ ∣∣tan(π4+x2)∣∣=−tan(π4+x2) (∴ in second quadrant)
=tan{π−(π4+x2)}
From Eq.(i),
y=tan−1tan{π−(π4+x2)}
=π−(π4+x2)
=3π4−x2
(∵principal value of tan−1 x in−π2 to π2)
∴ dydx=−12
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