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 ........

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Lời giải:

a)

\(\frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{(\sin ^2a+\cos ^2a)+\cos ^2a-1}{\cot ^2a}=\frac{1+\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{(\frac{\cos a}{\sin a})^2}=\sin ^2a\)

b)

\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)

\(=\frac{\sin ^2a+\cos ^2a}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\tan ^2a+1-1=\tan ^2a\)

c)

\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}=\frac{\sin ^4a(\cos ^2a-1)}{\cos ^4a(\sin ^2a-1)}\)

\(=\frac{\sin ^4a(-\sin ^2a)}{\cos ^4a(-\cos ^2a)}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

\(C=2sin3x.cos3x.\frac{cos3x}{sin3x}-\left(cos^23x-sin^23x\right)\)

\(=2cos^23x-cos^23x+sin^23x=cos^23x+sin^23x=1\)

\(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)\)

\(=\sqrt{2}\left(sinx.sin\frac{\pi}{4}-cosx.cos\frac{\pi}{4}\right)=-\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

Câu này bạn ghi nhầm đề (lưu ý rằng \(sin\frac{\pi}{4}=cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\))

Câu 2b bạn cũng xem lại đề, chắc chắn ko đúng

\(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^0\Rightarrow sin\frac{A}{2}=cos\left(\frac{B}{2}+\frac{C}{2}\right)=cos\frac{B}{2}cos\frac{C}{2}-sin\frac{B}{2}sin\frac{C}{2}\)

Câu 3 bạn cũng ghi sai đề luôn

Trong 1 ngày đẹp trời thì câu 4 cũng sai luôn cho đỡ lạc lõng đồng đội:

\(sin\left(a+b-b\right)=sin\left(a+b\right)cosb-cos\left(a+b\right)sinb=2sin\left(a+b\right)\)

\(\Leftrightarrow sin\left(a+b\right)\left[cosb-2\right]=cos\left(a+b\right).sinb\)

\(\Leftrightarrow\frac{sin\left(a+b\right)}{cos\left(a+b\right)}=\frac{sinb}{cosb-2}\Leftrightarrow tan\left(a+b\right)=\frac{sinb}{cosb-2}\)

4 câu bạn ghi đúng đề bài duy nhất câu 1, kinh thiệt :(