\(0< a< \frac{\pi}{2}\Rightarrow cosa>0\Rightarrow cosa=\sqrt{1-sin^2a}=\frac{4}{5}\)

\(\Rightarrow tana=\frac{sina}{cosa}=\frac{3}{4}\) ; \(cota=\frac{1}{tana}=\frac{4}{3}\)

\(\Rightarrow A=\frac{\frac{4}{3}+\frac{3}{4}}{\frac{4}{3}-\frac{3}{4}}=...\)

\(\frac{2sina+3cosa}{4sina-5cosa}=\frac{\frac{2sina}{cosa}+\frac{3cosa}{cosa}}{\frac{4sina}{cosa}-\frac{5cosa}{cosa}}=\frac{2tana+3}{4tana-5}=\frac{2.3+3}{4.3-5}=...\)

\(A=\frac{2sin^2a-3cos^2a}{sin^2a-2sina.cosa-cos^2a}=\frac{\frac{2sin^2a}{sin^2a}-\frac{3cos^2a}{sin^2a}}{\frac{sin^2a}{sin^2a}-\frac{2sina.cosa}{sin^2a}-\frac{cos^2a}{sin^2a}}=\frac{2-3cot^2a}{1-2cota-cot^2a}=\frac{2-3.3^2}{1-2.3-3^2}=...\)

\(A=\dfrac{4sina+5cosa}{2sina-3cosa}=\dfrac{\dfrac{4sina}{sina}+\dfrac{5cosa}{sina}}{\dfrac{2sina}{sina}-\dfrac{3cosa}{sina}}=\dfrac{4+5cota}{2-3cota}=\dfrac{4+5.\left(\dfrac{1}{2}\right)}{2-3.\left(\dfrac{1}{2}\right)}=...\)

\(sina+cosa=\sqrt{2}\Leftrightarrow\left(sina+cosa\right)^2=2\\ \)

\(\Leftrightarrow\sin^2a+2\sin a.cosa+cos^2a=2\)

\(\Leftrightarrow1+2.sina.cosa=2\)

\(\Leftrightarrow2.sina.cosa=2-1=1\)

\(\Leftrightarrow\sin a.cosa=\frac{1}{2}\)

Vậy  P=sina.cosa=\(\frac{1}{2}\)

\(Q=\sin^4a+cos^4a\)

\(\Leftrightarrow\left(sin^2a\right)^2+\left(cos^2a\right)^2\)

\(\Leftrightarrow\left(sin^2a+cos^2a\right)^2-2.sin^2a.cos^2a\)

\(\Leftrightarrow1^2-2.sin^2a.cos^2a\) tách tiếp rồi thế vào là được .tương tự phàn P ý
còn R thì tách sin^3a=sin^2a+sina tương tự cos mũ 3 a cụng vậy
theo tớ là như thế còn có sai thì đừng có ném đá ném gạch na

\(\frac{sin^2a+2sina.cosa-2cos^2a}{2sin^2a-3sina.cosa+4cos^2a}=\frac{\frac{sin^2a}{sin^2a}+\frac{2sina.cosa}{sin^2a}-\frac{2cos^2a}{sin^2a}}{\frac{2sin^2a}{sin^2a}-\frac{3sina.cosa}{sin^2a}+\frac{4cos^2a}{sin^2a}}\)

\(=\frac{1+2cota-2cot^2a}{2-3cota+4cot^2a}=\frac{1-6-2.9}{2+9+4.9}=...\)

\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)

\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)

Biết cotα=\(\dfrac{1}{2}\) nên ta có:

\(A=\dfrac{4+5\cdot\dfrac{1}{2}}{2-3\cdot\dfrac{1}{2}}\)

\(A=\dfrac{4+\dfrac{5}{2}}{2-\dfrac{3}{2}}\)

A= 13

cotx=2

=>cosx=2*sin x

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

=>\(\dfrac{1}{sin^2x}=1+4=5\)

=>\(sin^2x=\dfrac{1}{5}\)

\(B=\dfrac{sin^2x-2\cdot sinx\cdot2\cdot sinx-1}{5\cdot4sin^2x+sin^2x-3}=\dfrac{-3sin^2x-1}{21sin^2x-3}\)

\(=\dfrac{-\dfrac{3}{5}-1}{\dfrac{21}{5}-3}=-\dfrac{8}{5}:\dfrac{6}{5}=-\dfrac{4}{3}\)

\(cotx=2\Rightarrow tanx=\dfrac{1}{2}\)

\(B=\dfrac{sin^2x-2sinx.cosx-1}{5cos^2x+sin^2x-3}\)

\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-\dfrac{1}{cos^2x}}{5+tan^2x-\dfrac{3}{cos^2x}}\)

\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-1-tan^2x}{5+tan^2x-3-3tan^2x}\)

\(\Leftrightarrow B=\dfrac{-2tanx-1}{2-2tan^2x}\)

\(\Leftrightarrow B=\dfrac{-2.\dfrac{1}{2}-1}{2-2.\dfrac{1}{4}}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)

cot x=2>0

=>sin x và cosx cùng dấu

=>sinx*cosx>0

\(1+cot^2x=\dfrac{1}{sin^2x}=1+4=5\)

=>sin^2x=1/5

=>cos^2x=4/5

\(B=\dfrac{1}{5}-2\cdot sinx\cdot cosx-\dfrac{1}{5}\cdot\dfrac{4}{5}+\dfrac{1}{5}-3\)

\(=\dfrac{2}{5}-\dfrac{4}{25}-3-2\cdot\dfrac{1}{\sqrt{5}}\cdot\dfrac{2}{\sqrt{5}}\)

\(=\dfrac{10}{25}-\dfrac{4}{25}-\dfrac{75}{25}-2\cdot\dfrac{2}{5}=\dfrac{-69}{25}-\dfrac{4}{5}=\dfrac{-89}{25}\)

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

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

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

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