Question

How do you verify the identity $\csc 2\theta = \dfrac{{\csc \theta }}{{2\cos \theta }}$ ?

Answer 1

Hint : Here, you are given an equation which involves trigonometric ratios cosecant and cosine, a standard relation between these two quantities is given. What you need to do is use all the material, precisely, all the trigonometric properties and identities that involve cosecant and cosine or in addition to that, maybe other ratios as well and try to convert the left-hand side equal to right-hand side in order to verify the identity.
$\csc 2\theta = \dfrac{1}{{\sin 2\theta }} \\\ \csc 2\theta = \dfrac{1}{{2\sin \theta \cos \theta }} \;$

Complete step by step solution:
Let us consider the left-hand side, $\csc 2\theta$ . Cosecant of any angle can be written as the inverse of the sine of that angle, that is, cosecant of an angle $\theta$ is equal to one divided by the sine of $\theta$ . Mathematically, we have, $\csc \theta = \dfrac{1}{{\sin \theta }}$ . In our case, the angle is given to be twice of $\theta$ , that is $2\theta$ . So, let us put the angle in the above identity, we get, $\csc 2\theta = \dfrac{1}{{\sin 2\theta }}$ . Now, the property which gives you the expansion of sine of the sum of any two angles is $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ , where $A$ and $B$ are any two angles.
$2\theta$ can be written as $\theta + \theta$ , so we can consider $A = \theta$ and $B = \theta$ . Let us put these in the above property. $\sin \left( {\theta + \theta } \right) = \sin \theta \cos \theta + \cos \theta \sin \theta = 2\sin \theta \cos \theta$ . So, we have,
$\csc 2\theta = \dfrac{1}{{\sin 2\theta }} \\\ \csc 2\theta = \dfrac{1}{{2\sin \theta \cos \theta }} \\\$
Again, we will use the property which states that the cosecant of any angle can be written as the inverse of the sine of that angle, since we have $\dfrac{1}{{\sin \theta }}$ on the right-hand side of the above equation.
$\csc 2\theta = \dfrac{1}{{2\sin \theta \cos \theta }} \\\ \csc 2\theta = \dfrac{1}{{2\cos \theta }}\left( {\dfrac{1}{{\sin \theta }}} \right) \\\ \csc 2\theta = \dfrac{1}{{2\cos \theta }}\left( {\csc \theta } \right) \\\ \therefore \csc 2\theta = \dfrac{{\csc \theta }}{{2\cos \theta }} \\\$
As you can see, the left-hand side is equal to the right-hand side, we have proved the identity.
Hence proved $\csc 2\theta = \dfrac{{\csc \theta }}{{2\cos \theta }}$ , identity verified

Note : Here, we have used two properties and one trick. The two properties were, cosecant of any angle is equal to the inverse of sine of that angle, that is $\csc \theta = \dfrac{1}{{\sin \theta }}$ and the property which gives you the sine of sum of any two angle. The trick we used here was, we expressed $2\theta$ as $\theta + \theta$ which led us to use the above-mentioned property. You need to memorize all the properties and tricks used here.