Question

sec $(\pi -\theta )$ = -sec$\theta$

• A. True
• B. False

Answer 1

The correct option is A

True

Let θ be an angle in the standard position in the I quadrant.

Let its terminal sides cut the circle with center 0 and radius r. Let A (x, y) and A'(x', y') be the point of intersection of the terminal side of angles θ and π-θ respectively with the circle. Let B & B' be the projection of A and A' respectively in the x - axis.

Angle AOB' = $\pi -\theta$, Angle AOB = $\theta$

In$△$OAB & $△$OA'B'

OA = OA'(radius of circle)

$\angle OBA$ = $\angle O{B}^{\prime }{A}^{\prime }$ = ${90}^0$

$\angle A^{\prime }OB$ = $\angle AOB$ = $\theta$

So, OAB $\cong$ $△$OA'B'

X' = -x, Y'=y

sec$(\pi -\theta$) = $\frac{r}{x^{\prime }}$ = $\frac{r}{-x}$ = -sec$\theta$

sec$(\pi -\theta$) = -sec$\theta$