Question

Given that sec 0= -37/12, what is the value of cot 0, for pi/2 < 0 pi?

Answer 1

Answer:

  cot(θ) = -12/35

Step-by-step explanation:

  cot(θ) = -1/√(sec(θ)^2 -1) . . . . negative in the second quadrant

  = -1/√((-37/12)^2 -1) = -1/(35/12)

  cot(θ) = -12/35

Answer 2

Answer:

$-\frac{12}{35}$

Step-by-step explanation:

Given,

$sec\theta = -\frac{37}{12}$

We know that,

$sec \theta = \frac{H}{B}$

Where, H is the hypotenuse of a right triangle and B is its base,

By making diagram,

By applying pythagoras theorem,

The Perpendicular of the triangle,

$P=\sqrt{H^2-B^2}=\sqrt{37^2-12^2}=\sqrt{1369-144}=\sqrt{1225}=35$

Hence, $cot \theta = -\frac{B}{P}=-\frac{12}{35}$

( We took the negative sign because $\frac{\pi}{2}< \theta < \pi$ )