Question

cho tanx= - \(\dfrac{2}{3}\) tính A = \(\dfrac{3sin^2x-cos^2x}{2sin^2x}\)

cho cotx = \(\dfrac{3}{5}\) tính A = \(\dfrac{sin^2x-5cos^2x}{2cos^2x}\)

Answer 1

Lời giải:

a. 

\(A=\frac{3}{2}-2(\frac{\cos x}{\sin x})^2=\frac{3}{2}-2.(\frac{1}{\tan x})^2=\frac{3}{2}-\frac{1}{2}(\frac{-3}{2})^2=-3\)

b.

\(A=\frac{1}{2}(\frac{\sin x}{\cos x})^2-\frac{5}{2}=2(\frac{1}{\cot x})^2-\frac{5}{2}=2(\frac{5}{3})^2-\frac{5}{2}=\frac{55}{18}\)

Answer 2

a, \(A=\dfrac{3sin^2\left(x\right)-cos^2\left(x\right)}{2sin^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\dfrac{cos^2\left(x\right)}{sin^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\cdot\dfrac{1}{tan^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\cdot\left(-\dfrac{3}{2}\right)^2=-3\)

b, \(A=\dfrac{sin^2\left(x\right)-5cos^2\left(x\right)}{2cos^2\left(x\right)}=\dfrac{1}{2}\dfrac{sin^2\left(x\right)}{cos^2\left(x\right)}-\dfrac{5}{2}=\dfrac{1}{2}\cdot\dfrac{1}{cot^2\left(x\right)}-\dfrac{5}{2}=\dfrac{1}{2}\cdot\left(\dfrac{5}{3}\right)^2-\dfrac{5}{2}=\dfrac{55}{18}\)