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.