Question

How do you find a double angle formula for $\sec \left( {2x} \right)$ in terms of only $\csc (x)$ and $\sec (x)$

Answer 1

Here we just need to apply the formula according to our requirement of the trigonometric formula. We know that $\sec x = \dfrac{1}{{\cos x}}$ and $\csc x = \dfrac{1}{{\sin x}}$. In the same way we can write $\sec 2x = \dfrac{1}{{\cos 2x}}$. So we can apply the formula of $\cos (x + x) = \cos x\cos x - \sin x\sin x$ and get our required result.

Complete step-by-step answer:
Here we are given to find the formula of $\sec 2x$ in terms of $\sec x{\text{ and }}\csc x$
We know that secant is reciprocal of the cosine trigonometric function. Hence we can say:
$\sec 2x = \dfrac{1}{{\cos 2x}}$ $- - - - (1)$
Now we just need to find the value of $\cos 2x$ but all should be in the terms of $\sec x{\text{ and }}\csc x$
Now we can write $\cos 2x = \cos (x + x)$
We can apply the formula here which is as:
$\cos (A + B) = \cos A\cos B - \sin A\sin B$
If we compare $\cos (x + x){\text{ with }}\cos (A + B)$ we will get:
$A = x \\\ B = x \\\$
So after that we will get:
$\cos 2x = \cos (x + x) = \cos x\cos x - \sin x\sin x$
$\cos 2x = {\cos ^2}x - {\sin ^2}x$ $- - - - - - (2)$
Now we know that $\cos x = \dfrac{1}{{\sec x}}{\text{ and sin}}x = \dfrac{1}{{\csc x}}{\text{ }}$
Substituting these values in equation (2) we will get:
$\cos 2x = {\cos ^2}x - {\sin ^2}x$
$\cos 2x = \dfrac{1}{{{{\sec }^2}x}} - \dfrac{1}{{{{\csc }^2}x}}$
Now we can take the LCM and write the above equation as:
$\cos 2x = \dfrac{{{{\csc }^2}x - {{\sec }^2}x}}{{({{\sec }^2}x)({{\csc }^2}x)}}$
Now we have got the value of $\cos 2x$ all in the terms of $\sec x{\text{ and }}\csc x$ and these two are the only terms in which we are told to bring the whole equation of the trigonometric functions. Hence we now just need to put the value of the trigonometric function which is $\cos 2x$ in the equation (1) and get the value of $\sec 2x$ in term of the two trigonometric functions which are $\sec x{\text{ and }}\csc x$
So putting $\cos 2x = \dfrac{{{{\csc }^2}x - {{\sec }^2}x}}{{({{\sec }^2}x)({{\csc }^2}x)}}$ in equation (1) we get:
$\sec 2x = \dfrac{1}{{\dfrac{{{{\csc }^2}x - {{\sec }^2}x}}{{({{\sec }^2}x)({{\csc }^2}x)}}}} = \dfrac{{({{\sec }^2}x)({{\csc }^2}x)}}{{{{\csc }^2}x - {{\sec }^2}x}}$
Hence we have found the formula for the double angle $\sec \left( {2x} \right)$ in terms of only $\csc (x)$ and $\sec (x)$.

Note: Here in these types of problems where the student is asked to find the formula for one trigonometric function in terms of another trigonometric function, the student must only focus on how he can relate those two terms. For this he must be aware of all the general formula of the trigonometric functions like:
$\sin x = \dfrac{1}{{\csc x}} \\\ \cos x = \dfrac{1}{{\sec x}} \\\ \tan x = \dfrac{1}{{\cot x}} \\\$