TG EAMCET 2 May 2025 Shift 1 Paper

Give line is 5x−12y+6=0
Slope of this line is m1=‌

−5

−12

=‌

5

12

Since, line L is perpendicular to this line, the product of their slope is -1 . Let the slope of line L be mL.
Then, mL×‌

5

12

=−1
⇒‌‌mL=−‌

12

5

The equation of a line in normal form is x‌cos‌θ+ysin‌θ=P, where P is the distance from the origin to the line and θ is the angle made by the perpendicular from the origin.
We have P=2 unit, So, the equation of line is x‌cos‌θ+ysin‌θ=2
Slope, mL=‌

−cos‌θ

sin‌θ

=−cot‌θ
∴−cot‌θ=‌

−12

5

⇒cot‌θ=‌

12

5

For y-intercept, x=0 then ysin‌θ=2
⇒‌‌y=‌

2

sin‌θ

And since y-intercept is positive, so
‌‌

2

sin‌θ

>0
‌⇒sin‌θ>0
⇒cot‌θ=‌

12

5

>0,θ must be in first quadrant.
Now, tan‌θ=‌

1

cot‌θ

=‌

5

12

So, tan‌θ+cot‌θ=‌

5

12

+‌

12

5

=‌

25+144

60

=‌

169

60