Question

Prove that $\dfrac{{\cos ec\left( {90^\circ - x} \right) \cdot \sin \left( {180^\circ - x} \right) \cdot \cot \left( {360^\circ - x} \right)}}{{\sec \left( {180^\circ + x} \right) \cdot \tan \left( {90^\circ + x} \right) \cdot \sin \left( { - x} \right)}} = 1$

Answer 1

Hint : Here, to prove that $\dfrac{{\cos ec\left( {90^\circ - x} \right) \cdot \sin \left( {180^\circ - x} \right) \cdot \cot \left( {360^\circ - x} \right)}}{{\sec \left( {180^\circ + x} \right) \cdot \tan \left( {90^\circ + x} \right) \cdot \sin \left( { - x} \right)}} = 1$ , we will take the LHS of the equation and try to bring it equal to RHS. Now,
$\Rightarrow \cos ec\left( {90^\circ - x} \right) = \sec x$
$\Rightarrow \sin \left( {180^\circ - x} \right) = \sin x$
$\Rightarrow \cot \left( {360^\circ - x} \right) = - \cot x$
$\Rightarrow \sec \left( {180^\circ + x} \right) = - \sec x$
$\Rightarrow \tan \left( {90^\circ + x} \right) = - \cot x$
$\Rightarrow \sin \left( { - x} \right) = - \sin x$
Substituting all these values in LHS, we will get LHS = RHS.

Complete step-by-step answer :
In this question, we have to prove that
$\dfrac{{\cos ec\left( {90^\circ - x} \right) \cdot \sin \left( {180^\circ - x} \right) \cdot \cot \left( {360^\circ - x} \right)}}{{\sec \left( {180^\circ + x} \right) \cdot \tan \left( {90^\circ + x} \right) \cdot \sin \left( { - x} \right)}} = 1$ - - - - - - - - - - - - - - - - (1)
For proving this, let us take the LHS of the equation (1). Therefore,
$\Rightarrow LHS = \dfrac{{\cos ec\left( {90^\circ - x} \right) \cdot \sin \left( {180^\circ - x} \right) \cdot \cot \left( {360^\circ - x} \right)}}{{\sec \left( {180^\circ + x} \right) \cdot \tan \left( {90^\circ + x} \right) \cdot \sin \left( { - x} \right)}}$
Now, let us find the value of each term separately.
$\cos ec\left( {90^\circ - x} \right)$ :
As we know, cosec and sec are complementary to each other, we have a relation between them that
$\Rightarrow \cos ec\left( {90^\circ - x} \right) = \sec x \\\ \Rightarrow \sec \left( {90^\circ - x} \right) = \cos ecx \;$
Therefore, $\cos ec\left( {90^\circ - x} \right) = \sec x$ - - - - - - (2)

$\sin \left( {180^\circ - x} \right)$ :
As x is being subtracted from 180, $\sin \left( {180^\circ - x} \right)$ lies between $\pi$ and $\dfrac{\pi }{2}$ that is the 2nd quadrant. And the sine and cosecant functions are positive in the 2nd quadrant. So, we get
$\Rightarrow \sin \left( {180^\circ - x} \right) = \sin x$ - - - - - - (3)

$\cot \left( {360^\circ - x} \right)$ :
As, x is being subtracted from 360, $\cot \left( {360^\circ - x} \right)$ lies between $2\pi$ and $\dfrac{{3\pi }}{2}$ that is 4th quadrant. And the cot function is negative in the 4th quadrant. So, we get
$\Rightarrow \cot \left( {360^\circ - x} \right) = - \cot x$ - - - - - - (4)

$\sec \left( {180^\circ + x} \right)$ :
As, we are adding x to 180, $\sec \left( {180^\circ + x} \right)$ will lie between $\pi$ and $\dfrac{{3\pi }}{2}$ that is 3rd quadrant. And in the 3rd quadrant, sec is negative. So, we get
$\Rightarrow \sec \left( {180^\circ + x} \right) = - \sec x$ - - - - - - (5)

$\tan \left( {90^\circ + x} \right)$ :
Now, here as we are adding x into 90, $\tan \left( {90^\circ + x} \right)$ will lie between $\pi$ and $\dfrac{\pi }{2}$ that is 2nd quadrant. And as the angle is $\dfrac{\pi }{2}$ , we have to change the function from tan to cot. Here, cot will be negative in the 2nd quadrant. So, we get
$\Rightarrow \tan \left( {90^\circ + x} \right) = - \cot x$ - - - - - - (6)

$\sin \left( { - x} \right)$ :
Now, we have a relation that
$\Rightarrow \sin \left( { - \theta } \right) = - \sin \theta$
Therefore,
$\Rightarrow \sin \left( { - x} \right) = - \sin x$ - - - - - - (7)

So, now putting the values of equations (2), (3), (4), (5), (6), (7) in equation (1), we get
$\Rightarrow LHS = \dfrac{{\cos ec\left( {90^\circ - x} \right) \cdot \sin \left( {180^\circ - x} \right) \cdot \cot \left( {360^\circ - x} \right)}}{{\sec \left( {180^\circ + x} \right) \cdot \tan \left( {90^\circ + x} \right) \cdot \sin \left( { - x} \right)}}$
$= \dfrac{{\sec x \cdot \sin x \cdot \left( { - \cot x} \right)}}{{ - \sec x \cdot \left( { - \cot x} \right) \cdot \left( { - \sin x} \right)}} \\\ = \dfrac{1}{{\left( { - 1} \right)\left( { - 1} \right)}} \\\ = 1 \;$
Hence, LHS = RHS.
Therefore, we proved that $\dfrac{{\cos ec\left( {90^\circ - x} \right) \cdot \sin \left( {180^\circ - x} \right) \cdot \cot \left( {360^\circ - x} \right)}}{{\sec \left( {180^\circ + x} \right) \cdot \tan \left( {90^\circ + x} \right) \cdot \sin \left( { - x} \right)}} = 1$ .

Note : Here, note that when the angle is made with X –Axis, we don’t change the trigonometric function and just change the sign according to the quadrant. But, when the angle is being made between X – Axis and Y – Axis, then we need to change the sign according to the quadrant and the trigonometric function as well.