Question

cotx = 5/12 then sinx + 1/cotx + secx

Answer 1

Answer:

the answer is given as follows

Answer 2

Answer

77/13

$\rule{200}1$

Explanation

Given,

cotx = 5/12

We know that cot x = B/P

Where, B is the base of the triangle (the side adjacent to angle x) and P is the perpendicular (the side opposite to the angle x)

The ratio of B/P = 5/12

So, let B = 5k and P = 12k

Now, by Pythagoras theorem, we know that

Hypotenuse = $\sf\sqrt{P^2 + B^2}$

Hence, Hypotenuse = $\sf\sqrt{(5k)^2 + (12k)^2}$

→ Hypotenuse = $\sf\sqrt{25k^2 + 144k^2}$

→ Hypotenuse = $\sf\sqrt{169k^2}$

→ Hypotenuse = 13k

Now, we know that sin∅ = P/H (H is the hypotenuse)

→ sin(x) = 12k/13k = 12/13

sec∅ = H/B

→ sec(x) = 13k/5k = 13/5

and, 1/cot(x) = 12/5

So,

sin(x) + 1/cot(x) + sec(x)

→ 12/13 + 12/5 + 13/5

→ 12/13 + 25/5

→ 12/13 + 5

→ 12/13 + 65/13

→ 77/13