Question

prove that (sec 8θ -1 ) / (sec4θ -1) = (tan8θ) / (tan 2θ)​

Answer 1

Given

$\\ \colon{\tt\red{ \dfrac{sec8 \theta -1 }{sec4 \theta -1} = \dfrac{tan8 \theta }{tan2 \theta } }}$

Solution:

$\\ \colon\implies{\tt{ \dfrac{sec8 \theta -1 }{sec4 \theta -1} = \dfrac{tan8 \theta }{tan2 \theta } }} \\ \\ \\ \colon\implies{\tt{ \dfrac{ \dfrac{1}{cos8 \theta} -1}{ \dfrac{1}{cos4 \theta} -1} = \dfrac{tan8 \theta }{tan2 \theta } }} \\ \\ \\ \colon\implies{\tt{ \dfrac{[1 - cos8 \theta]cos4 \theta }{[1 - cos4 \theta]cos8 \theta} = \dfrac{tan8 \theta }{tan2 \theta } }} \\ \\ \\ \colon\implies{\tt{ \dfrac{ 2sin^2 4 \theta cos 4 \theta }{ [2sin^2 2 \theta ] cos8 \theta} = \dfrac{tan8 \theta }{tan2 \theta } }} \\ \\ \\ \colon\implies{\tt{ \dfrac{ 2sin \ 4 \theta \ cos \ 4 \theta \ sin \ 4 \theta }{ [2sin^2 \ 2 \theta ] cos \ 8 \theta} = \dfrac{tan8 \theta }{tan2 \theta } }} \\ \\ \\ \colon\implies{\tt{ \dfrac{ sin \ 8 \theta · 2sin2 \theta cos2 \theta }{[2sin^2 \ 2 \theta ] cos \ 8 \theta } = \dfrac{tan8 \theta }{tan2 \theta } }} \\ \\ \\ \colon\implies{\tt{ \dfrac{tan \ 8 \theta · cos2 \theta }{sin 2 \theta} = \dfrac{tan8 \theta }{tan2 \theta } }} \\ \\ \\ \colon\implies{\tt{ \dfrac{tan8 \theta }{tan2 \theta } = \dfrac{tan8 \theta }{tan2 \theta } }} \\ \\ \\ \colon{\tt\large\green{LHS = RHS }}$

Hence Proved !!!