Question

Class 11th
Prove that

$\frac{sec8a-1}{sec4a-1} = \frac{tan8a}{tan2a}$

Answer 1

Multiply the left hand side by a special kind of 1 : $\frac{\sec8a+1}{\sec8a+1}$

$\frac{\sec8a-1}{\sec4a-1}$

$=\frac{\sec8a-1}{\sec4a-1}\times\frac{\sec8a+1}{\sec8a+1}$

$=\frac{\sec^28a-1}{(\sec4a-1)(\sec8a+1)}~~~\because(a-b)(a+b)=a^2-b^2$

$=\frac{\tan^28a}{\frac{(1-\cos4a)(1+\cos8a)}{\cos4a\cos8a}}~~~\because\sec^2\theta-1=\tan^2\theta$

$=\frac{\tan^28a\cos4a\cos8a}{(2\sin^22a)(2\cos^24a)}~~~\because1-\cos2\theta=2\sin^2\theta,~1+\cos2\theta=2\cos^2\theta$

$=\frac{\tan8a\sin8a}{4\sin^22a\cos4a}$

$=\frac{\tan8a~2\sin4a}{4\sin^22a}$

$=\frac{\tan8a~4\cos2a}{4\sin2a}$

$=\frac{\tan8a}{\tan2a}$

Answer 2

 

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plz solve dis problem

prove that:


(sec8A - 1)/(sec4A - 1) = tan8A/tan2A

7 years ago

Answers : (5)

(SEC8A -1)/SEC4A-1 = TAN8A/TAN2A

SECX =1/COSX

SO

    LHS = (1-COS8A)COS4A/(1-COS4A)COS8A

    COS8A=1-2SIN24A    

    COS4A =1-2SIN22A

AFTER PUTTING THESE

  LHS = (SIN24A)COS4A/(SIN22A)COS8A

        = (2SIN4ACOS4A)SIN4A/2SIN22ACOS8A                             (multiplying dividing by 2)

2SIN4ACOS4A =COS8A

 SO

    LHS =(SIN8A)SIN4A/(2SIN22A)COS8A =(TAN8A)SIN4A/2SIN22A

 SIN4A = 2SIN2ACOS2A

SO

  LHS =(TAN8A)COS2A/SIN2A=TAN8A/TAN2A = RHS

 HENCE PROVED