\(\frac{sinA}{cosA}+\frac{sinB}{cosB}=\frac{2cos\frac{C}{2}}{sin\frac{C}{2}}\Leftrightarrow\frac{sinA.cosB+cosA.sinB}{cosA.cosB}=\frac{2sin\frac{C}{2}.cos\frac{C}{2}}{sin^2\frac{C}{2}}\)

\(\Leftrightarrow\frac{sin\left(A+B\right)}{cosA.cosB}=\frac{2sinC}{1-cosC}\Leftrightarrow\frac{sinC}{cosA.cosB}=\frac{2sinC}{1-cosC}\)

\(\Leftrightarrow1-cosC=2cosA.cosB=cos\left(A+B\right)+cos\left(A-B\right)\)

\(\Leftrightarrow1-cosC=-cosC+cos\left(A-B\right)\)

\(\Leftrightarrow cos\left(A-B\right)=1\Rightarrow A-B=0\Rightarrow A=B\)

\(\Rightarrow\) Tam giác ABC cân tại C

\(\frac{cos^2A+cos^2B}{sin^2A+sin^2B}=\frac{1}{2}\left(cot^2A+cot^2B\right)\)

\(\Leftrightarrow2cos^2A+2cos^2B=\left(sin^2A+sin^2B\right)\left(cot^2A+cot^2B\right)\)

\(\Leftrightarrow2cos^2A+2cos^2B=cos^2A+cos^2B+sin^2A.cot^2B+sin^2B.cot^2A\)

\(\Leftrightarrow cos^2A+cos^2B=\frac{sin^2A.cos^2B}{sin^2B}+\frac{sin^2B.cos^2A}{sin^2A}\)

\(\Leftrightarrow cos^2A\left(\frac{sin^2B}{sin^2A}-1\right)=cos^2B\left(1-\frac{sin^2A}{sin^2B}\right)\)

\(\Leftrightarrow\frac{cos^2A\left(sin^2B-sin^2A\right)}{sin^2A}=\frac{cos^2B\left(sin^2B-sin^2A\right)}{sin^2B}\)

\(\Leftrightarrow cot^2A\left(sin^2B-sin^2A\right)=cot^2B\left(sin^2B-sin^2A\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2B=sin^2A\\cot^2A=cot^2B\end{matrix}\right.\) \(\Rightarrow A=B\)

Gỉa sử \(\Delta ABC\)cân tại C, kẻ \(CH⊥AB\)

Ta có VT= \(\cos^2A=\frac{AH^2}{AC^2};\cos^2B=\frac{BH^2}{BC^2}\Rightarrow\cos^2A+\cos^2B=\frac{AH^2}{AC^2}+\frac{BH^2}{BC^2}=2.\frac{AH^2}{AC^2}\)do \(\hept{\begin{cases}AH=BH\\AC=BC\end{cases}}\)

\(\sin^2A=\frac{CH^2}{CA^2};\sin^2B=\frac{CH^2}{CB^2}\Rightarrow\sin^2A+\sin^2B=2.\frac{CH^2}{CA^2}\)

\(\Rightarrow\frac{\cos^2A+\cos^2B}{\sin^2A+\sin^2B}=\frac{2.\frac{AH^2}{AC^2}}{2.\frac{CH^2}{AC^2}}=\frac{AH^2}{CH^2}\)

Ta có VP =\(\frac{1}{2}\left(\cot^2A+\cot^2B\right)=\frac{1}{2}.\left(\frac{AH^2}{CH^2}+\frac{BH^2}{CH^2}\right)=\frac{1}{2}\left(2.\frac{AH^2}{CH^2}\right)=\frac{AH^2}{CH^2}\)

Ta thấy VT=VP\(\Rightarrow\)giả sử đúng 

Vậy ........

\(\frac{cos\left(a-b\right)}{sin\left(a+b\right)}=\frac{cosa.cosb+sina.sinb}{sina.cosb+cosa.sinb}=\frac{\frac{cosa.cosb}{sina.sinb}+1}{\frac{sina.cosb}{sina.sinb}+\frac{cosa.sinb}{sina.sinb}}=\frac{cota.cotb+1}{cota+cotb}\)

Bạn ghi đề ko đúng

\(sin\left(a+b\right)sin\left(a-b\right)=\frac{1}{2}\left[cos2b-cos2a\right]\)

\(=\frac{1}{2}\left[1-2sin^2b-1+2sin^2a\right]\)

\(=sin^2a-sin^2b\)

\(=1-cos^2a-1+cos^2b=cos^2b-cos^2a\)

Câu này bạn cũng ghi đề ko đúng

\(cos\left(a+b\right)cos\left(a-b\right)=\frac{1}{2}\left[cos2a+cos2b\right]\)

\(=\frac{1}{2}\left[2cos^2a-1+1-2sin^2b\right]=cos^2a-sin^2b\)

\(=1-sin^2a-1+cos^2b=cos^2b-sin^2a\)

MN K BT?