AP EAPCET 18 May 2024 Shift 1 Solved Paper

To determine ‌

dy

dx

, we begin with the expressions given for x and y.
x=3[sin‌t−log(cot‌

t

2

)]
Differentiating x with respect to t :
‌

dx

dt

‌=3[cos‌t−‌

1

cot‌ t 2

⋅(−csc2‌

t

2

)⋅‌

1

2

]
‌=3[cos‌t+‌

1

sin‌t

]
‌=3[‌

cos‌t‌s‌i‌n‌t+1

sin‌t

]
Next, differentiating y with respect to t :
y=‌6[cos‌t+log(tan‌

t

2

)]
‌

dy

dt

‌=6[−sin‌t+‌

1

tan‌ t 2

⋅ sec2‌

t

2

⋅‌

1

2

]
‌=6[−sin‌t+‌

1

sin‌t

]
‌=6[‌

−sin‌2t+1

sin‌t

]
Now, to find ‌

dy

dx

:
‌

dy

dx

=‌

‌ dy dt

dx dt

=‌

6[ −sin‌2t+1 sin‌t ]

3[‌ cos‌t‌s‌i‌n‌t+1 sin‌t ]

The sin‌t terms cancel out:
=‌

6(−sin‌2t+1)

3(cos‌t‌s‌i‌n‌t+1)

Simplifying further:
=‌

2(−sin‌2t+1)

cos‌t‌s‌i‌n‌t+1

Rewriting the numerator, 1−sin‌2t=cos2t :
=‌

2cos2t

cos‌t‌s‌i‌n‌t+1

Therefore, the expression for ‌

dy

dx

is:
‌

dy

dx

=‌

2cos2t

1+sin‌t‌cos‌t