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

Biết tanα=\(-\dfrac{1}{4}\) nên ta có:

\(\dfrac{3\cdot\dfrac{-1}{4}-4}{2\cdot\dfrac{-1}{4}+3}=\dfrac{-\dfrac{3}{4}-4}{-\dfrac{1}{2}+3}=\dfrac{-19}{10}\)

\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)

\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)

a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)

b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)

Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)

\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)

Dạ em cảm ơn thầy và mọi người ạ! 

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

\(=cos^2a.sin^2a-cos^2a=cos^2a\left(sin^2a-1\right)=-cos^4a\)

\(B=\sqrt{\left(1-cos^2a\right)^2+6cos^2a+3cos^4a}+\sqrt{\left(1-sin^2a\right)^2+6sin^2a+3sin^4a}\)

\(=\sqrt{4cos^4a+4cos^2a+1}+\sqrt{4sin^4a+4sin^2a+1}\)

\(=\sqrt{\left(2cos^2a+1\right)^2}+\sqrt{\left(2sin^2a+1\right)^2}\)

\(=2\left(sin^2a+cos^2a\right)+2=4\)

1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)

\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)

\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)

\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)

1: \(cota=\sqrt{5}\)

=>\(cosa=\sqrt{5}\cdot sina\)

\(1+cot^2a=\dfrac{1}{sin^2a}\)

=>\(\dfrac{1}{sin^2a}=1+5=6\)

=>\(sin^2a=\dfrac{1}{6}\)

\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)

\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)

2: tan a=3

=>sin a=3*cosa 

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)

\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)

\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)

\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)

A

Chọn A

b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)

hay \(\cos\alpha=\dfrac{4}{5}\)

Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)

\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)

\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)

\(=\dfrac{141}{25}\)

c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)

\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)

\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)

\(sin\alpha=sin\left(180-\alpha\right)=\dfrac{3}{5}\Rightarrow cos\left(180-a\right)=\sqrt{1-sin^2\alpha}=\dfrac{4}{5}\Rightarrow cos\alpha=-\dfrac{4}{5}\)

\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{5}}{-\dfrac{4}{5}}=-\dfrac{3}{4}\Rightarrow cot\alpha=-\dfrac{4}{3}\)

\(\Rightarrow A=\dfrac{3.\dfrac{3}{5}-\dfrac{4}{5}}{-\dfrac{3}{4}+\dfrac{4}{3}}=\dfrac{12}{7}\)

a, \(\dfrac{1-sin2a}{1+sin2a}\)

\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)

\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)

\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)

\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)

\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)

b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)

\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)

\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)

\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)

\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)

\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)

Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)

Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)

\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)

\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)

\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)

\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)