Question

If tanA=a then find sin4A-2sin2A/ sin4A+2sin2A​

Answer 1

Answer:

The value of the given expression is

$\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2$

Step by step Explanation:

Given that $tanA=a$

To find the value of $\frac{sin4A-2sin2A}{sin4A+2sin2A}$

Let us find the value of the expression :

$\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A}$ ( by using the formula $sin2A=2cosAsinA$ here A=2A)  

$=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}$

$=\frac{(cos2A-1)}{(cos2A+1)}$

$=\frac{(-(1-cos2A))}{(1+cos2A)}$(using  $sin^2A+cos^2A=1$  here A=2A)

$=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}$(using $cos2A=cos^2A-sin^2A$ here A=2A)

$=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}$

$=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}$

$=\frac{-2sin^2A}{2cos^2A}$

$=-\frac{sin^2A}{cos^2A}$

$=-tan^2A$  ( using $tanA=\frac{sinA}{cosA}$ here A=2A )

 $=-a^2$ (since tanA=a given )

Therefore $\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2$