Question

Find the value of the following expression:
$\dfrac{{\cos (360 - A)}}{{\sin (270 + A)}} + \dfrac{{\cot (90 + A)}}{{\tan (180 - A)}} + \dfrac{{\sin (90 - A)}}{{\sin (90 + A)}}$

Answer 1

Hint:We know that in the first quadrant$(0 < \theta < 90)$, all are positive. In the second quadrant $(90 < \theta < 180)$ only sin and cos are positive and in the third quadrant$(180 < \theta < 270)$, only tan and cot are positive and in the fourth quadrant$(270 < \theta < 360)$, only cos and sec are positive and we also know that $\sin (90 \pm \theta ) = \pm \cos \theta $ and $\sin (180 \pm \theta ) = \pm \sin \theta $ and so on.Using this concept we try to solve the question.

Complete step-by-step answer:
Here we should know in which quadrant, which trigonometric function is positive or negative which means that where sin, cos, tan, cot, sec and cosec are positive or negative.

In the first quadrant$(0 < \theta < 90)$, all are positive. In the second quadrant $(90 < \theta < 180)$ only sin and cos are positive and in the third quadrant$(180 < \theta < 270)$, only tan and cot are positive and in the fourth quadrant$(270 < \theta < 360)$, only cos and sec are positive.
Here we also know that
$\sin (90 \pm \theta ) = \pm \cos \theta $
$\sin (180 \pm \theta ) = \pm \sin \theta $
$\cos (90 \pm \theta ) = \pm \sin \theta $
$\tan (90 \pm \theta ) = \pm \cot \theta $
$\sec (90 \pm \theta ) = \pm \cos ec\theta $
Where $\theta $ is the acute angle.
Here we are given:
$\dfrac{{\cos (360 - A)}}{{\sin (270 + A)}} + \dfrac{{\cot (90 + A)}}{{\tan (180 - A)}} + \dfrac{{\sin (90 - A)}}{{\sin (90 + A)}}$
Here $\cos (360 - A)$ is given and we know that $(360 - A)$ lies in the fourth quadrant therefore the cos will be positive.
$\cos $$(360 - A)$$ = \cos A$$ - - - - - (1)$
Also we can say that $(270 + A)$ lies in the fourth quadrant so
$\sin (270 + A) = - \cos A$$ - - - - (2)$
$\cot (90 + A) = - \tan A$
$\tan (180 - A) = - \tan A$$ - - - - (3)$
As we know that $\sin (90 - A) = \cos A$
90+A lies in the second quadrant and sin is positive so
$\sin (90 + A) = \cos A$$ - - - (4)$
So using the values from the above equations in the given problem, we get:
$\dfrac{{\cos (360 - A)}}{{\sin (270 + A)}} + \dfrac{{\cot (90 + A)}}{{\tan (180 - A)}} + \dfrac{{\sin (90 - A)}}{{\sin (90 + A)}}$
$\dfrac{{\cos A}}{{ - \cos A}} + \dfrac{{-\tan A}}{{ - \tan A}} + \dfrac{{\cos A}}{{\cos A}}$
$ - 1 + 1 + 1 = 1$
So the value of expression is $1$
Note:Students should remember trigonometric formulas i.e.$\sin \theta .\cos ec\theta = 1,\tan \theta .\cot \theta = 1$ and $\cos \theta .\sec \theta = 1$ and also identities ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , $1+{\tan^2}\theta = {\sec^2}\theta$ and $1+{\cot^2}\theta = {\cos sec^2}\theta$ .Also we have to remember that in which quadrant, which trigonometric function is positive or negative which means that where sin, cos, tan, cot, sec and cosec are positive or negative.