Question

Prove that \[\dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A}}{{\tan 2A}}\]

Answer 1

Hint: Here, we are asked to prove the given expression. To prove this take one side of the given equation that is (left hand side) L.H.S or (right hand side) R.H.S and then simplify it using the trigonometry formulas. If you take the L.H.S then try to simplify it to make it equal to R.H.S and if you take R.H.S then try to make it equal to L.H.S. While solving, try to use the trigonometric formulas to simplify the steps.

Complete step-by-step answer:
We are asked to prove the equation \[\dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A}}{{\tan 2A}}\]
First let us take the L.H.S \[ = \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}}\]
We know, \[\sec x = \dfrac{1}{{\cos x}}\]
Using this on L.H.S we get,
\[{\text{L}}{\text{.H}}{\text{.S}} = \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} \\
   = \dfrac{{\dfrac{1}{{\cos 8A}} - 1}}{{\dfrac{1}{{\cos 4A}} - 1}} \\
= \dfrac{{\left( {1 - \cos 8A} \right)\cos 4A}}{{\left( {1 - \cos 4A} \right)\cos 8A}}\] (i)
We can write, \[{\sin ^2}x = \dfrac{1}{2}(1 - \cos 2x)\]
\[ \Rightarrow 2{\sin ^2}x = 1 - \cos 2x\]
Using this in equation (i) we get
\[{\text{L}}{\text{.H}}{\text{.S}} = \dfrac{{\left( {2{{\sin }^2}4A} \right)\left( {\cos 4A} \right)}}{{\left( {2{{\sin }^2}2A} \right)\left( {\cos 8A} \right)}}\]
\[ = \dfrac{{\left( {2\sin 4A\cos 4A} \right)\sin 4A}}{{\left( {2{{\sin }^2}2A} \right)\left( {\cos 8A} \right)}}\] (ii)
We can write \[2\sin x\cos x = \sin 2x\], using this formula in equation (ii), we get
\[ \dfrac{{\left( {2\sin 4A\cos 4A} \right)\sin 4A}}{{\left( {2{{\sin }^2}2A} \right)\left( {\cos 8A} \right)}} \\
   = \dfrac{{\sin 8A\sin 4A}}{{\left( {\cos 8A} \right)\left( {2{{\sin }^2}2A} \right)}} \\
   = \left( {\dfrac{{\sin 8A}}{{\cos 8A}}} \right)\left( {\dfrac{{\sin 4A}}{{2{{\sin }^2}2A}}} \right) \\
   = \tan 8A\left( {\dfrac{{\sin 4A}}{{2{{\sin }^2}2A}}} \right) \]
Again using the formula \[2\sin x\cos x = \sin 2x\] for \[\sin 4A\], we get
\[ \tan 8A\left( {\dfrac{{\sin 4A}}{{2{{\sin }^2}2A}}} \right) \\
   = \tan 8A\left( {\dfrac{{2\sin 2A\cos 2A}}{{2{{\sin }^2}2A}}} \right) \\
   = \tan 8A\left( {\dfrac{{\cos 2A}}{{\sin 2A}}} \right) \\
   = \dfrac{{\tan 8A}}{{\tan 2A}} \]

Therefore, we get
\[{\text{L}}{\text{.H}}{\text{.S}} = \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A}}{{\tan 2A}}\]


Note: Whenever you are asked to prove an equation, always start with one side of the equation and then try to simplify it in such a way that it is equal to the other side of the equation. For trigonometry questions, always remember the trigonometric identities and formulas. In such types of questions, try to convert the given trigonometric functions into sine or cosine functions as it makes the steps easy and also there are many formulas for sine and cosine functions which we can use to get our desired result. Like here the function given was sec and we know sec is inverse of cosine so we converted it to cosec, similarly try to solve for other questions too.