Chọn C

\(A=sin42^0-cos48^0=cos\left(90^0-42^0\right)-cos48^0=cos48^0-cos48^0=0\)

\(B=cot56^0-tan34^0=tan\left(90^0-56^0\right)-tan34^0=tan34^0-tan34^0=0\)

\(C=sin30^0-cot50^0-cos60^0+tan40^0\)

\(=cos\left(90^0-30^0\right)-tan\left(90^0-50^0\right)-cos60^0+tan40^0\)

\(=cos60^0-tan40^0-cos60^0+tan40^0=0\)

\(A=\sin42^0-\cos48^0=\sin42^0-\sin42^0=0\)

\(B=\cot56^0-\tan34^0=\tan34^0-\tan34^0=0\)

đáp án

A=Sin 42^o - cos 48^o =cos(90^o - 42^o) - cos 48^o= cos48^o - cos48^o=0

hok tốt

B=cos56^o-tan34^o=tan(90^o - 56^o) - tan34^o=tan34^o - tan34^o=0

a: Ta có: \(A=\left(\sin30^0-cos60^0\right)+\left(\tan40^0\cdot\tan50^0\right)\)

\(=\left(\sin30^0-\sin30^0\right)+\left(\tan40^0\cdot\cot40^0\right)\)

=0+1

=1

b: \(B=\left(cos^220^0+cos^270^0\right)-\left(\cot42^0\cdot\cot48^0\right)\)

\(=\left(\sin^220^0+cos^220^0\right)-\left(\cot42^0\cdot\tan42^0\right)\)

=1-1

=0

Ta có: \(\frac{\tan\alpha}{\cot\alpha}+\frac{\cot\alpha}{\tan\alpha}-\frac{\sin^2\alpha}{cos^2\alpha}\)

\(=1+1-\tan^2\alpha=2-\tan^2\alpha\)

A

Chọn A

a: \(\frac{3\cdot\cot60^0}{2\cdot cos^230^0-1}\)

\(=\frac{3\cdot\frac{1}{\sqrt3}}{2\cdot\left(\frac{\sqrt3}{2}\right)^2-1}=\frac{\sqrt3}{2\cdot\frac34-1}=\frac{\sqrt3}{\frac32-1}=\sqrt3:\frac12=2\sqrt3\)

b: \(\frac{cos60^0}{1+\sin60^0}+\frac{1}{\tan30^0}\)

\(=\frac12:\left(1+\frac{\sqrt3}{2}\right)+1:\frac{1}{\sqrt3}=\frac12:\frac{2+\sqrt3}{2}+\sqrt3=\frac{1}{2+\sqrt3}+\sqrt3\)

\(=2-\sqrt3+\sqrt3\)

=2

a)1-1+1=1

a.1

b.1,140119483

c.0,353338015