Math, asked by Shum2187, 10 months ago

Prove that: Sec8theta - 1 / sec4theta - 1 = tan8theta / tan2theta

Answers

Answered by wwwmohdsabirc
5

Step-by-step explanation:

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Answered by sandy1816
0

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

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