Question

TÍNH

a) A= tan 1 độ* tan 2 độ * tan 3 độ.....tan 89 độ

b) Cho góc nhọn α,tan α=\(\dfrac{1}{2}\) tính:

B=\(\dfrac{\sin\alpha+2\cos\alpha}{3\sin\alpha-4\cos\alpha}\)

D=\(\dfrac{2\sin^2\alpha-3\cos^2\alpha}{4\cos^2\alpha-5\sin^2\alpha}\)

GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP

Answer 1

a) ta có : \(A=tan1.tan2.tan3...tan89\)

\(=\left(tan1.tan89\right).\left(tan2.tan88\right).\left(tan3.tan87\right)...\left(tan44.tan46\right).tan45\)

\(=\left(tan1.tan\left(90-1\right)\right).\left(tan2.tan\left(90-2\right)\right).\left(tan3.tan\left(90-3\right)\right)...\left(tan44.tan\left(90-44\right)\right).tan45\)

\(=\left(tan1.cot1\right).\left(tan2.cot2\right).\left(tan3.cot3\right)...\left(tan44.cot44\right).tan45\) \(=tan45=1\)

b) ta có \(B=\dfrac{sin\alpha+2cos\alpha}{3sin\alpha-4cos\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}\)

\(=\dfrac{tan\alpha+2}{3tan\alpha-4}=\dfrac{\dfrac{1}{2}+2}{\dfrac{3}{2}-4}=-1\)

ta có \(D=\dfrac{2sin^2\alpha-3cos^2\alpha}{4cos^2\alpha-5sin^2\alpha}=\dfrac{\dfrac{2sin^2\alpha}{cos^2\alpha}-\dfrac{3cos^2\alpha}{cos^2\alpha}}{\dfrac{4cos^2\alpha}{cos^2\alpha}-\dfrac{5sin^2\alpha}{cos^2\alpha}}\)

\(=\dfrac{2tan^2\alpha-3}{4-5tan^2\alpha}=\dfrac{2\left(\dfrac{1}{2}\right)^2-3}{4-5\left(\dfrac{1}{2}\right)^2}=\dfrac{-10}{11}\)