Question

Find the value of expression
$\sin (-\theta )+\cos (-\theta )+\sec (-\theta )$
$+\sin (\pi -\theta )+\cos (\pi -\theta )+\sec (\pi -\theta )$

• A. $2\sin \theta$
• B. $2\cos \theta$
• C. $2\sec \theta$
• D. 0

Answer 1

The correct option is D

0

Solution: First let's find the value of $\sin (-\theta )$,

Let $\theta$ be an angle in the standard position in the I quadrant.

$A(x,y)$ and $A^{\prime }(x^{\prime },y^{\prime })$ be the point of intersection of the terminal side of angle $-\theta$ in the standard position with the circle.

Now $\angle BOA=\angle BO{A}^{\prime }$ (numerically) and $A$ & $A^{\prime }$ have the same projection $B$ on the $x-$axis.

$\angle OBA=$$\angle OB{A}^{\prime }=$${90}^{\circ }$

$OA=O{A}^{\prime }$ (radius of circle)

$△BOA$ $\cong$ $△BO{A}^{\prime }$

$x=x^{\prime }$ & $y=-y^{\prime }$

$\sin (-\theta )=$ $\frac{y^{\prime }}{r}$ = $\frac{-y}{r}=$$-\sin \theta$

Similarly, $\cos (-\theta )=$$\cos \theta$ & $\sec (-\theta$) $=\sec \theta$

Let $\theta$ be an angle in the standard position in the $1^{st}$ quadrant.

Let its terminal sides cut the circle with center 0 and radius r. Let $A(x,y)$ and $A^{\prime }(x^{\prime },y^{\prime })$ be the

point of intersection of the terminal side of angles $\theta$ and $\pi -$$\theta$ respectively with the circle. Let $B$ &

$B^{\prime }$ be the projection of $A$ and $A^{\prime }$ respectively in the $x-$ axis.

In $△OAB$ & $△O{A}^{\prime }{B}^{\prime }$

$OA=O{A}^{\prime }$ radius of circle

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

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

So, $OAB\cong$ $△O{A}^{\prime }{B}^{\prime }$

$X^{\prime }=-x,Y^{\prime }=y$

$\sin (\pi$ $-\theta$) = $\frac{y^{\prime }}{r}=$ $\frac{y}{r}=$ $\sin \theta$
$\sin (\pi -\theta )$$=\sin \theta$

Similarly, $\cos (\pi -\theta )$ $=-\cos \theta$ & $\sec (\pi -\theta )$ $=-\sec \theta$

From the given expression

$\sin (-\theta )+\cos (-\theta )+\sec (-\theta )+\sin (\pi -\theta )+\cos (\pi -\theta )+\sec (\pi -\theta )$

$-\sin \theta +\cos \theta +\sec \theta +\sin \theta -\cos \theta -\sec \theta =0$