The slope of the normal to the curve y=sin^2x at is:
a) 1
b)-1
c)
Answers
Answered by
2
Answer:
b)-1
Step-by-step explanation:
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Answered by
2
Answer:
We have,
y=(sin2x+cotx+2)
2
On differentiating both sides w.r.t. x, we get,
dx
dy
=2(sin2x+cotx+2)
dx
d
(sin2x+cotx+2)
dx
dy
=2(sin2x+cotx+2)(2cos2x−cosec
2
x)
Therefore, (
dx
dy
)
x=π/2
=sinπ+cot
2
π
+2
dx
dy
=2(sin2x+cotx+2)(2cos2x−cosec
2
x)
Therefore, (
dx
dy
)
x=π/2
=sinπ+cot
2
π
+2
=2[0+0+2][−2−1]=−12
Therefore, slope of tangent = -12
And, slope of normal =
12
1
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