Question

X=log tan t , y=log sin t

Answer 1

Answer:

We are given, x=a(cost+logtan

2

t

)

and y=asint

Let differentiate x & y wrt t, we get,

dt

dy

=acost

&

dt

dx

=a

⎣

⎢

⎢

⎡

(−sint)+

tan

2

t

1

.sec

2

2

t

.

2

1

⎦

⎥

⎥

⎤

( by chain rule)

dt

dx

=a

⎣

⎢

⎢

⎡

−sint+

2sin

2

t

1cos

2

t

×

cos

2

2

t

1

⎦

⎥

⎥

⎤

=a

⎣

⎢

⎢

⎡

−sint+

2sin

2

t

cos

2

t

1

⎦

⎥

⎥

⎤

=a[−sint+

sint

1

]

dt

dx

=

sint

a(1−sin

2

t

=

sint

acos

2

t

now,

dx

dy

=

dt

dy

×

dx

dt

=

acos

2

t

acost

×sint=tant

again differentiate

dx

dy

wrt x.

dx

d

(

dx

dy

)=

dx

2

d

2

y

=sec

2

t.

dx

dt

dx

2

d

2

y

=sec

2

t×

acos

2

t

sint

[

dx

2

d

2

y

]

t=

3

π

=

a.cos

2

(π/3)

sec

2

(π/3).sin(π/3)

=

a.(1/2)

2

(2)

2

.(

3

/2)

[

dx

2

d

2

y

]

t=

3

π

=

a

8

3

Step-by-step explanation:

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