Question

Prove
\[\dfrac{\cos\left( 2\pi\ + \ x \right){cosec}\left( 2\pi\ + \ x \right)\tan\left( \left( \dfrac{\pi}{2} \right) + \ x \right)}{\sec\left( \left( \dfrac{\pi}{2} \right) + \ x \right){cos xcot}\left( \pi\ + \ x \right)} = 1\]

Answer 1

Hint: In this question, we have to prove that
\[\dfrac{\cos\left( 2\pi\ + \ x \right){cosec}\left( 2\pi\ + \ x \right)\tan\left( \left( \dfrac{\pi}{2} \right) + \ x \right)}{\sec\left( \left( \dfrac{\pi}{2} \right) + \ x \right){cos xcot}\left( \pi\ + \ x \right)}\] is equal to \[1\] . First we need to consider the left hand side of the given expression. In order to prove this,we need to use the concepts of trigonometric identities. By using trigonometric identities and functions, we can prove this expression easily.

Complete answer:
Consider the left part of the expression,
$\Rightarrow$ \[\dfrac{\cos\left( 2\pi\ + \ x \right){cosec}\left( 2\pi\ + \ x \right)\tan\left( \left( \dfrac{\pi}{2} \right) + \ x \right)}{\sec\left( \left( \dfrac{\pi}{2} \right) + \ x \right){cos xcot}\left( \pi\ + \ x \right)}\] ••• (1)
We know that the value of \[\pi\] is \[180°\]
$\Rightarrow$ \[2\pi = 2(180°)\]
By multiplying,
We get,
\[2\pi = 360°\]
We can rewrite \[cos(2\pi + x)\] as \[{cosx}\], since it lies in the first quadrant.
Similarly, we can also rewrite \[cosec(2\pi + x)\] as \[{cosec x}\], since it lies in the first quadrant.
Now we need to find \[tan(\dfrac{\pi}{2} + x)\],
We know that \[tanx = \dfrac{{sinx}}{{cosx}}\]
\(tan(\dfrac{\pi}{2} + x)\ = \dfrac{\sin\left( \dfrac{\pi}{2} + x \right)}{\cos\left( \left( \dfrac{\pi}{2} \right) + x \right)}\ \)
We also know that,
\[sin(A + B)\ = sinAcosB + cosAsinB\]
\[cos(A + B)\ = cosAcosB – sinAsinB\]
Thus we get,
\[tan(\dfrac{\pi}{2} + x)\ = \dfrac{\sin\left( \frac{\pi}{2} \right)cosx + \cos\left( \dfrac{\pi}{2} \right){sinx}}{\cos\left( \dfrac{\pi}{2} \right)cosx - \sin\left( \dfrac{\pi}{2} \right){sinx}}\ \]
We know that,
\[s{in}\left( \dfrac{\pi}{2} \right) = 1\]
\[c{os}\left( \dfrac{\pi}{2} \right) = 0\]
By substituting the values,
We get,
$\Rightarrow$ \[\tan\left( \left( \frac{\pi}{2} \right) + x \right)\ = \frac{1 \times cosx + 0}{0 – 1 \times sinx}\]
\[= \dfrac{{cosx}}{- sinx}\]
We know that \[\dfrac{{cosx}}{{sinx}}\] is \[{cotx}\]
Thus we get,
$\Rightarrow$ \[\tan\left( \left( \frac{\pi}{2} \right) + x \right) = - cotx\]
Similarly, we need to find for
\[\sec\left( \left( \dfrac{\pi}{2} \right) + x \right)\]
$\Rightarrow$ \[\sec\left( \left( \dfrac{\pi}{2} \right) + x \right) = \dfrac{1}{\cos\left( \left( \dfrac{\pi}{2} \right) + x \right)}\]
We know that \[cos(\left( \dfrac{\pi}{2} \right) + x)\] is \[\ - sinx\]
Thus we get,
\[\sec\left( \left( \dfrac{\pi}{2} \right) + x \right) = \dfrac{1}{- sinx}\]
We also know that \[\dfrac{1}{\sin x}\ \] is \[{cosecx}\]
Thus we get,
\[\sec\left( \left( \dfrac{\pi}{2} \right) + x \right) = - cosecx\]
Then by rewriting the terms in (1),
We get,
$\Rightarrow$ \[\dfrac{{cosxcosecx}\left( - cotx \right)}{\left( - cosecx \right){cosxcotx}}\]
By simplifying,
We get,
\[\dfrac{\cos\left( 2\pi\ + \ x \right){cosec}\left( 2\pi\ + \ x \right)\tan\left( \left( \dfrac{\pi}{2} \right) + \ x \right)}{\sec\left( \left( \dfrac{\pi}{2} \right) + \ x \right){cos xcot}\left( \pi\ + \ x \right)} = 1\]
Thus we have proved that,
\[\dfrac{\cos\left( 2\pi\ + \ x \right){cosec}\left( 2\pi\ + \ x \right)\tan\left( \left( \dfrac{\pi}{2} \right) + \ x \right)}{\sec\left( \left( \dfrac{\pi}{2} \right) + \ x \right){cos xcot}\left( \pi\ + \ x \right)} = 1\]
Final answer :
\[\dfrac{\cos\left( 2\pi\ + \ x \right){cosec}\left( 2\pi\ + \ x \right)\tan\left( \left( \dfrac{\pi}{2} \right) + \ x \right)}{\sec\left( \left( \dfrac{\pi}{2} \right) + \ x \right){cos xcot}\left( \pi\ + \ x \right)} = 1\]

Note:
The concept used to prove the given expression is trigonometric identities. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the substitution rule with the use of trigonometric functions.