Question

Biểu thức \(\dfrac{2cos^2a-1}{4tan\left(\dfrac{\pi}{4}-a\right)sin^2\left(\dfrac{\pi}{4}+a\right)}\) có kết quả rút gọn bằng?

Answer 1

\(=\dfrac{2\cdot\cos^2a-1}{4\cdot\dfrac{1-tana}{1+tana}\cdot\dfrac{1}{2}\left(\cos a+\sin a\right)^2}\)

\(=\dfrac{2\cdot\cos^2a-1}{2\cdot\left(cosa+sina\right)^2\cdot\dfrac{\cos a-sina}{cosa+sina}}\)

\(=\dfrac{2\cdot cos^2a-1}{2\cdot\left(cosa+sina\right)\left(cosa-sina\right)}\)

\(=\dfrac{2\left(1-sin^2a\right)-1}{2\cdot\left(cos^2a-sin^2a\right)}\)

\(=\dfrac{2-2sin^2a-1}{2\cdot\left(cos^2a-sin^2a\right)}=\dfrac{-2sin^2a+1}{2\left(cos^2a-sin^2a\right)}\)