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# Prove that $\dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A}}{{\tan 2A}}$

Last updated date: 12th Aug 2024
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Answer
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Hint: Use conversion of secant function into cosine function. Use the trigonometric identity $\cos 2x = 1 - 2{\sin ^2}x$ to open the values required in the numerator and denominator. Combine the terms to form suitable pairs using the identity $2\sin x\cos x = \sin 2x$ in the end.
* $\sec x = \dfrac{1}{{\cos x}}$
* $\cos 2x = 1 - 2{\sin ^2}x$
* $\tan x = \dfrac{{\sin x}}{{\cos x}}$

Complete step-by-step solution:
We have to prove $\dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A}}{{\tan 2A}}$
We solve the left hand side of the equation
Since LHS is $\dfrac{{\sec 8A - 1}}{{\sec 4A - 1}}$, we transform secant functions to cosine functions using $\sec x = \dfrac{1}{{\cos x}}$
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\dfrac{1}{{\cos 8A}} - 1}}{{\dfrac{1}{{\cos 4A}} - 1}}$
Take LCM in both numerator and denominator of RHS of the equation
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\dfrac{{1 - \cos 8A}}{{\cos 8A}}}}{{\dfrac{{1 - \cos 4A}}{{\cos 4A}}}}$
Write the right hand side of the equation in simpler form
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\left( {1 - \cos 8A} \right)}}{{\cos 8A}} \times \dfrac{{\cos 4A}}{{\left( {1 - \cos 4A} \right)}}$,,,,,,,,,,,,,,,,,,,,,,, … (1)
Now we calculate the values of $\cos 8A$and$\cos 4A$using the identity $\cos 2x = 1 - 2{\sin ^2}x$
Since, $\cos 2x = 1 - 2{\sin ^2}x$
$\Rightarrow \cos 4A = \cos 2(2A) = 1 - 2{\sin ^2}(2A)$ …………………...… (2)
Similarly,
$\Rightarrow \cos 8A = \cos 2(4A) = 1 - 2{\sin ^2}(4A)$..................… (3)
Substitute the values from equations (2) and (3) in brackets of equation (1)
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\left( {1 - \left( {1 - 2{{\sin }^2}(4A)} \right)} \right)}}{{\cos 8A}} \times \dfrac{{\cos 4A}}{{\left( {1 - \left( {1 - 2{{\sin }^2}(2A)} \right)} \right)}}$
Calculate the value inside the brackets in both numerator and denominator
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\left( {1 - 1 + 2{{\sin }^2}(4A)} \right)}}{{\cos 8A}} \times \dfrac{{\cos 4A}}{{\left( {1 - 1 + 2{{\sin }^2}(2A)} \right)}}$
Cancel same terms having opposite signs inside the bracket
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\left( {2{{\sin }^2}(4A)} \right)}}{{\cos 8A}} \times \dfrac{{\cos 4A}}{{\left( {2{{\sin }^2}(2A)} \right)}}$
Write numerator and denominators after multiplication
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\left( {2{{\sin }^2}(4A)} \right)\cos 4A}}{{\left( {2{{\sin }^2}(2A)} \right)\cos 8A}}$.....................… (4)
We can write $\left( {2{{\sin }^2}(4A)} \right)\cos 4A = \left( {2\sin 4A\cos 4A} \right)\sin 4A$
Use the identity$2\sin x\cos x = \sin 2x$ to write$\left( {2\sin 4A\cos 4A} \right) = \sin 2(4A) = \sin 8A$
$\Rightarrow \left( {2{{\sin }^2}(4A)} \right)\cos 4A = \sin 8A\sin 4A$
Equation (4) becomes
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\sin 8A\sin 4A}}{{\left( {2{{\sin }^2}(2A)} \right)\cos 8A}}$
Since $\dfrac{{\sin x}}{{\cos x}} = \tan x \Rightarrow \dfrac{{\sin 8A}}{{\cos 8A}} = \tan 8A$
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A\sin 4A}}{{\left( {2{{\sin }^2}(2A)} \right)}}$................… (5)
Now we can write $\sin 4A = 2\sin 2A\cos 2A$ using the identity $2\sin x\cos x = \sin 2x$
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A2\sin 2A\cos 2A}}{{2\sin 2A\sin 2A}}$
Cancel same factors from numerator and denominator in right hand side of the equation
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A\cos 2A}}{{\sin 2A}}$
Write $\dfrac{{\sin 2A}}{{\cos 2A}} = \tan 2A$
$\Rightarrow \dfrac{{\sec 8A - 1}}{{\sec 4A - 1}} = \dfrac{{\tan 8A}}{{\tan 2A}}$
This is equal to the right hand side of the given equation.
$\therefore$LHS $=$RHS
Hence Proved

Note: Students are likely to make mistake of substituting all the values of $\cos 4A$ and $\cos 8A$ in both numerator and denominator, keep in mind the right hand side of the equation has angles of tangent as 8A and 2A, so don’t change all the trigonometric terms in the equation, only change the part that seems that it can be reduced to smaller form.