Answer
See the full explanation below.
Work Step by Step
If the identities are provided in $\sin x$ and $\cos x$ then, there is no change in the identities.
Let us take an example:
The identities are:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
$\cos x=\sqrt{1-{{\sin }^{2}}x}$
$\sin x=\sqrt{1-{{\cos }^{2}}x}$
If the identity is provided in $\csc x$, $\sec x$ $\tan x$ and $\cot x$, then, we convert the identities into the form of $\sin x$ and $\cos x$ respectively as shown below,
For $\csc x$
$\csc x=\frac{1}{\sin x}$
For $\sec x$,
$\sec x=\frac{1}{\cos x}$
For $\tan x$, $\tan x=\frac{\sin x}{\cos x}$
For $\cot x$
$\cot x=\frac{\cos x}{\sin x}$
Then, the steps to verify the identities are.
Step 1: consider the complicated part of the identity. The complicated part may be the left-hand side or the right-hand part.
Step 2: convert the identity into $\sin x$ or $\cos x$ as per the provided identity.
Step 3: simplify the complicated part until we get the same identity as another part.
For example:
Now, consider an identity,
$\tan \theta +\cot \theta =\left( \csc \theta \right)\left( \sec \theta \right)$
Consider the left-hand side and simplify:
$\begin{align}
& L.H.S.=\tan \theta +\cot \theta \\
& =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \\
& =\frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta }
\end{align}$
Since, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ then,
$\begin{align}
& L.H.S.=\frac{1}{\sin \theta \cos \theta } \\
& =\csc \theta \sec \theta
\end{align}$
Hence, from the above calculation, the left part of the identity is equal to the right part of the identity.
