a: \(\sin^2a+cos^2a=1\)

=>\(cos^2a=1-\left(\frac45\right)^2=1-\frac{16}{25}=\frac{9}{25}\)

=>\(cosa=\frac35\)

\(A=5\cdot\sin a+3\cdot cosa\)

\(=5\cdot\frac45+3\cdot\frac35=4+\frac95=\frac{29}{5}\)

b:Ta có: \(\tan a\cdot\cot a=1\)

=>\(\tan a=1:\frac13=3\)

\(B=\frac{\sin a-cosa}{\sin a+cosa}\)

\(=\frac{\sin a-cosa}{cosa}:\frac{\sin a+cosa}{cosa}=\frac{\left(\tan a-1\right)}{\tan a+1}\)

\(=\frac{3-1}{3+1}=\frac24=\frac12\)

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

= \(\frac{sin\alpha}{cos\alpha}:\frac{cos\alpha}{sin\alpha}+\frac{cos\alpha}{sin\alpha}:\frac{sin\alpha}{cos\alpha}-\frac{sin^2\alpha}{cos^2\alpha}\)

= \(\frac{cos^2\alpha}{sin^2\alpha}=cot^2\alpha\)

Bạn gõ lại đề dc ko chứ mk đọc đề bài của bạn mà lú luôn............

      \(2\sin30^0-2cos60^0+\tan45^0\)

\(=2cos30^0-2cos30^0+1=1\)(Do trong tam giác có 2 góc phụ nhau thì sin góc này bằng cos góc kia)

1.

\(\frac{\pi}{2}< x< \pi\\ \Rightarrow cosx< 0,sinx>0,cotx< 0\)

\(cotx=\frac{1}{tanx}=\frac{-1}{3}\)

\(1+tan^2x=\frac{1}{cos^2x}\\ \Rightarrow cosx=\sqrt{\frac{1}{1+tan^2}}=\sqrt{\frac{1}{1+9}}=-\frac{\sqrt{10}}{10}\)

\(sinx=\sqrt{1-cos^2x}=\sqrt{1-\frac{10}{100}}=\frac{3\sqrt{10}}{10}\)

ta có : \(B=\left(1+tan^2x\right)\left(1-sin^2x\right)-\left(1+cot^2x\right)\left(1-cos^2x\right)\)

\(=\left(1+\dfrac{sin^2x}{cos^2x}\right)\left(sin^2x+cos^2x-sin^2x\right)-\left(1+\dfrac{cos^2x}{sin^2x}\right)\left(sin^2x+cos^2x-cos^2x\right)\)

\(=\dfrac{sin^2x+cos^2x}{cos^2x}\left(cos^2x\right)-\dfrac{sin^2x+cos^2x}{sin^2x}\left(sin^2x\right)\)

\(=\dfrac{1}{cos^2x}.cos^2x-\dfrac{1}{sin^2x}.\left(sin^2\right)x=1-1=0\)

nhớ ghi góc nha bn :) .