Question

(Sinx+cosx)/(sinx-cosx) =3, then the value of sin^4x-cos^4x is

Answer 1

Answer:

$\text{The value of sin^4x-cos^4x is \frac{3}5{} }$

Step-by-step explanation:

$\textbf{Given:}$

$\frac{sinx+cosx}{sinx-cosx} =3$

$\implies\,sinx+cosx=3(sinx-cosx)$

$\implies\,sin\,x+cos\,x=3\,sin\,x-3\,cos\,x$

$\implies\,4\,cos\,x=2\,sin\,x$

$\implies\,2\,cos\,x=sin\,x$

$\implies\,tan\,x=2$

$\text{Now,}$

$sin^4x-cos^4x$

$=cos^4x(\frac{sin^4x}{cos^4x}-1)$

$=\frac{1}{sec^4x}(tan^4x-1)$

$=\frac{1}{(sec^2x)^2}(tan^4x-1)$

$=\frac{1}{(1+tan^2x)^2}(tan^4x-1)$

$=\frac{1}{(1+2^2)^2}(2^4-1)$

$=\frac{1}{5^2}(2^4-1)$

$=\frac{1}{25}(15)$

$=\frac{15}{25}$

$=\frac{3}{5}$

$\implies\boxed{\bf\,sin^4x-cos^4x=\frac{3}{5}}$