Question

\[\dfrac{cosec\left( 90{}^\circ -x \right)sin\left( 180{}^\circ -x \right)cot\left( 360{}^\circ -x \right)}{sec\left( 180{}^\circ +x \right)tan\left( 90{}^\circ +x \right)sin\left( -x \right)}=1\]

Answer 1

Hint: In this question, we have to prove the given equation. The equation contains quite a few trigonometric identities which could not be used as it is. Neither could we use the value of these trigonometric functions since exact angles are known. So, in order to prove that the left hand side is equal to the right hand side, we will simplify each trigonometric function using trigonometric identities. We can also derive these identities while doing the solution. We use trigonometric ratios for various angles of the form (90°-x), (90°+x) and so on.

Complete step by step solution :
We have to prove the equation
\[\dfrac{cosec\left( 90{}^\circ -x \right)sin\left( 180{}^\circ -x \right)cot\left( 360{}^\circ -x \right)}{sec\left( 180{}^\circ +x \right)tan\left( 90{}^\circ +x \right)sin\left( -x \right)}=1\]
The equation involves trigonometric identities
In order to prove this we will consider from LHS and then equate it to RHS .
\[LHS=\dfrac{cosec\left( 90{}^\circ -x \right)sin\left( 180{}^\circ -x \right)cot\left( 360{}^\circ -x \right)}{sec\left( 180{}^\circ +x \right)tan\left( 90{}^\circ +x \right)sin\left( -x \right)}\]
Now using trigonometric identities , we know that
\[cosec\left( 90{}^\circ -x \right)=\dfrac{1}{sin\left( 90{}^\circ -x \right)}\]
And using trigonometric identity for \[sin\left( 90{}^\circ -x \right)=cosx\]
Therefore
\[cosec\left( 90{}^\circ -x \right)=\dfrac{1}{cosx}=secx\]
And \[sin\left( 180{}^\circ -x \right)=sin\left( 90{}^\circ -\left( 90{}^\circ -x \right) \right)\]
                         \[\begin{align}
  & =\cos\left( 90{}^\circ -x \right) \\
 & ~=\sin \text{ }x \\
\end{align}\]
(Since \[\cos\left( 90{}^\circ -x \right)=\sin x\])
\[\begin{align}
  & \begin{array}{*{35}{l}}
   cot\left( 360{}^\circ -x \right)=-cotx \\
   sec\left( 180{}^\circ +x \right)=-secx \\
   tan\text{ }\left( 90{}^\circ +x \right)=-cotx \\
\end{array} \\
 & sin\left( -x \right)=-sinx \\
\end{align}\]
Now we will substitute all the trigonometric functions with their values .
On substituting , we get
\[\begin{align}
  & LHS=\dfrac{cosec\left( 90{}^\circ -x \right)sin\left( 180{}^\circ -x \right)cot\left( 360{}^\circ -x \right)}{sec\left( 180{}^\circ +x \right)tan\left( 90{}^\circ +x \right)sin\left( -x \right)} \\
 & =\dfrac{\sec x\cdot \sin x\cdot (-\cot x)}{(-\sec x)\cdot (-\cot x)\cdot (-\sin x)} \\
\end{align}\]
secx , sinx and cotx gets cancelled from both numerator and denominator
Hence we get
\[\begin{align}
  & LHS=\dfrac{\sec x\cdot \sin x\cdot (-\cot x)}{(-\sec x)\cdot (-\cot x)\cdot (-\sin x)} \\
 & =\dfrac{-1}{-1} \\
 & =1 \\
 & =RHS \\
\end{align}\]
Hence proved.

Note :
It is important to note that we can derive all the trigonometric ratios for all angles using the value for sin, cos and tan but for convenience, it is always better to memorize the values for other trigonometric ratios for all types of angles as \[\left( 90-x \right)\text{ },\text{ }\left( 90+x \right)\text{ },\text{ }\left( 180-x \right)\text{ },\text{ }\left( 180+x \right)\text{ },\text{ }\left( 270-x \right)\text{ },\text{ }\left( 270+x \right)\text{ },\text{ }\left( 360-x \right)\] and \[\left( 360+x \right)\]. Another point to be noted is to take care of all the signs in the trigonometric identities.