Question

Prove that $\frac{sec8\theta -1}{sec4\theta -1}=\frac{tan8\theta }{tan2\theta }$

Answer 1

LHS = $\frac{sec8\theta -1}{sec4\theta -1}=\frac{\frac{1}{cos8\theta }-1}{\frac{1}{cos4\theta }-1}=\frac{\frac{1-cos8\theta }{cos8\theta }}{\frac{1-cos8\theta }{cos4\theta }}=\left(\frac{1-cos8\theta }{1-cos4\theta }\right)\left(\frac{cos4\theta }{cos8\theta }\right)\left[1\right]$

= $\frac{2si{n}^{2}4\theta }{2si{n}^{2}2\theta }.\frac{cos4\theta }{cos8\theta }=\frac{2sin4\theta .cos4\theta .sin4\theta }{2si{n}^{2}2\theta .cos8\theta }[\because 1-cos2A=2si{n}^{2}a]\left[1\right]$

= $\frac{sin8\theta .sin4\theta }{cos8\theta .2si{n}^{2}2\theta }=\frac{sin8\theta .2sin2\theta .cos2\theta }{cos8\theta .2si{n}^{2}2\theta }$

= $\frac{sin8\theta }{cos8\theta }.\frac{cos2\theta }{sin2\theta }=\frac{tan8\theta }{tan2\theta }=RHS[\because sin2A=2sinAcosA]\left[2\right]$