Question

Solve the following:
 $\dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} - \dfrac{{\tan 8A}}{{\tan 4A}}$

Answer 1

Hint: There are two terms in the above problem. Simplify the first term separately and then try to bring it in closer form to the second term. Check if they are equal or have any relation.

Complete step by step answer:
Our problem is: \[\dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} - \dfrac{{\tan 8A}}{{\tan 4A}}\]
We will consider the first term, rearrange it with cosine terms and see if we come closer to the second term.
$
\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} \\
\Rightarrow \dfrac{{\dfrac{1}{{\cos 8A}} - 1}}{{\dfrac{1}{{\cos 4A}} - 1}} = \dfrac{{1 - \cos 8A}}{{1 - \cos 4A}} \times \dfrac{{\cos 4A}}{{\cos 8A}} \\
$
But we know that, $\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = 1 - 2{\sin ^2}\alpha $
We will use this for the first term of the product in the last step. We will get,
$
\Rightarrow \dfrac{{1 - (1 - 2{{\sin }^2}4A)}}{{1 - (1 - 2{{\sin }^2}2A)}} \times \dfrac{{\cos 4A}}{{\cos 8A}} \\
\Rightarrow \dfrac{{{{\sin }^2}4A}}{{{{\sin }^2}2A}} \times \dfrac{{\cos 4A}}{{\cos 8A}} \\
$
We also know that, $\sin 2\alpha = 2\sin \alpha \cos \alpha $. We will use this relation to club together sine and cosine terms in the numerator. After multiplying both numerator and denominator by 2, we get,
$
\Rightarrow \dfrac{{2\sin 4A\cos 4A \times \sin 4A}}{{2{{\sin }^2}2A \times \cos 8A}} \\
\Rightarrow \dfrac{{\sin 8A \times \sin 4A}}{{2{{\sin }^2}2A \times \cos 8A}} \\
$
Therefore, we get,
$
\Rightarrow \dfrac{{\sin 8A}}{{\cos 8A}} \times \dfrac{{\sin 4A}}{{2{{\sin }^2}2A}} \\
\Rightarrow \tan 8A \times \dfrac{{2\sin 2A\cos 2A}}{{2{{\sin }^2}2A}} \\
\Rightarrow \tan 8A \times \dfrac{1}{{\tan 2A}} \\
$
At least we have reached a step where there are some terms in common between the first and second terms in our problem.
$ \Rightarrow \tan 8A\left[ {\dfrac{1}{{\tan 2A}} - \dfrac{1}{{\tan 4A}}} \right]$
But we know that, $\tan 2\alpha = \dfrac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}$. We will use this for the second term in brackets. We will get,
$
\Rightarrow \tan 8A \times \left[ {\dfrac{1}{{\tan 2A}} - \dfrac{{1 - {{\tan }^2}2A}}{{2\tan 2A}}} \right] \\
\Rightarrow \tan 8A \times \left[ {\dfrac{{2 - (1 - {{\tan }^2}2A)}}{{2\tan 2A}}} \right] \\
\Rightarrow \tan 8A \times \left[ {\dfrac{{1 + {{\tan }^2}2A}}{{2\tan 2A}}} \right] \\
\Rightarrow \tan 8A \times \left[ {\dfrac{{1 + \dfrac{{{{\sin }^2}2A}}{{{{\cos }^2}2A}}}}{{2\dfrac{{\sin 2A}}{{\cos 2A}}}}} \right] \\
\Rightarrow \tan 8A \times \left[ {\dfrac{{\dfrac{{{{\cos }^2}2A + {{\sin }^2}2A}}{{{{\cos }^2}2A}}}}{{2\dfrac{{\sin 2A}}{{\cos 2A}}}}} \right] \\
$
We know that, ${\cos ^2}\alpha + {\sin ^2}\alpha = 1$. We will use this relation in the ultimate numerator inside the bracket and cancel common terms to get,
$
\Rightarrow \tan 8A \times \left[ {\dfrac{1}{{2\sin 2A\cos 2A}}} \right] \\
\Rightarrow \dfrac{{\tan 8A}}{{\sin 4A}} \\
$. Here we have used the identity used in earlier steps.
Any further attempt will keep on adding terms and not simplify it. At $A = 0$, above solution goes to undefined values. At $A = \dfrac{\pi }{8}$, we get, 0. We see that there are steep changes in the values and no unique values for individual values of A.

Note: For such problems, it is better to simplify the terms using different identities of trigonometry and try to explain the range of values the final solution varies between for different values of the independent variable here A.