Question

prove that sec 8theta - 1/sec 4theta - 1 = tan 8theta/tan 2 theta

Answer 1

★ TRIGONOMETRIC EQUATIONS ★

Answer 2

Answer:

LHS

$\frac{sec 8\theta - 1}{sec4 \theta - 1} \\ \\ = \frac{(1 -cos8 \theta)cos4 \theta }{(1 - cos 4\theta)cos 8\theta} \\ \\ = \frac{(1 - 1 + 2 {sin}^{2} 4 \theta)cos4 \theta}{(1 - 1 + 2 {sin}^{2}2 \theta)cos8 \theta } \\ \\ = \frac{2 {sin}^{2} 4 \theta cos4 \theta}{2 {sin}^{2} \theta cos8 \theta }$

$= \frac{2sin4 \theta cos4 \theta sin4 \theta}{2sin2 \theta cos8 \theta sin2 \theta} \\ \\ = \frac{sin8 \theta2sin2 \theta cos2 \theta}{2sin2 \theta cos8 \theta sin2 \theta} \\ \\ = \frac{tan8 \theta}{tan2 \theta}$

RHS