b: \(\frac{\tan a+\tan b}{\cot a+cotb}\)

\(=\frac{\tan a+\tan b}{\frac{1}{\tan a}+\frac{1}{\tan b}}=\frac{\tan a+\tan b}{\frac{\tan a+\tan b}{\tan a\cdot\tan b}}=\tan a\cdot\tan b\)

c: \(\frac{\tan^2a-\tan^2b}{\tan^2a\cdot\tan^2b}\)

\(=\left(\frac{\sin^2a}{cos^2a}-\frac{\sin^2b}{cos^2b}\right):\left(\frac{\sin^2a}{cos^2a}\cdot\frac{\sin^2b}{cos^2b}\right)\)

\(=\frac{\left(\sin^2a\cdot cos^2b\right)-\sin^2b\cdot cos^2a}{cos^2a\cdot cos^2b}:\frac{\sin^2a\cdot\sin^2b}{cos^2a\cdot cos^2b}\)

\(=\frac{\left(\sin a\cdot cosb-sinb\cdot cosa\right)\left(\sin a\cdot cosb+\sin b\cdot cosa\right)}{\sin^2a\cdot\sin^2b}\)

\(=\frac{\sin\left(a-b\right)\cdot\sin\left(a+b\right)}{\sin^2a\cdot\sin^2b}=\frac{\frac12\cdot\left\lbrack cos\left(a-b-a-b\right)-cos\left(a-b+a+b\right)\right\rbrack}{\sin^2a\cdot sin^2b}\)

\(=\frac{\frac12\cdot\left\lbrack cos\left(-2b\right)-cos2a\right\rbrack}{\sin^2a\cdot\sin^2b}=\frac{\frac12\cdot\left\lbrack cos2b-cos2a\right\rbrack}{\sin^2a\cdot\sin^2b}=\frac{\frac12\cdot\left\lbrack1-2\cdot\sin^2b-1+2\cdot\sin^2a\right\rbrack}{\sin^2a\cdot\sin^2b}\)

\(=\frac{\sin^2a-\sin^2b}{\sin^2a\cdot\sin^2b}\)

a: \(\sin\left(a+b\right)\cdot\sin\left(a-b\right)\)

\(=\frac12\cdot\left\lbrack cos\left(a+b-a+b\right)-cos\left(a+b+a-b\right)\right\rbrack\)

\(=\frac12\left\lbrack cos2b-cos2a\right\rbrack=\frac12\cdot\left\lbrack2\cdot cos^2b-1-\left(2\cdot cos^2a-1\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack2\cdot cos^2b-2\cdot cos^2a\right\rbrack=cos^2b-cos^2a\)

\(=\left(1-\sin^2b\right)-\left(1-\sin^2a\right)=\sin^2a-\sin^2b\)

b: \(4\cdot\sin\left(x+\frac{\pi}{3}\right)\cdot\sin\left(x-\frac{\pi}{3}\right)\)

\(=4\cdot\frac12\cdot\left\lbrack cos\left(x+\frac{\pi}{3}-x+\frac{\pi}{3}\right)-cos\left(x+\frac{\pi}{3}+x+\frac{\pi}{3}\right)\right\rbrack\)

\(=2\cdot\left\lbrack cos\left(\frac23\pi\right)-cos2x\right\rbrack=2\cdot\left\lbrack-\frac12-cos2x\right\rbrack=-1-2\cdot cos2x\)

\(=-1-2\cdot\left(1-2\cdot\sin^2x\right)=-1-2+4\cdot\sin^2x=4\cdot\sin^2x-3\)

c: \(\sin\left(x+\frac{\pi}{4}\right)-\sin\left(x-\frac{\pi}{4}\right)\)

\(=\sin x\cdot cos\left(\frac{\pi}{4}\right)+cosx\cdot\sin\left(\frac{\pi}{4}\right)-\left\lbrack\sin x\cdot cos\left(\frac{\pi}{4}\right)-cosx\cdot\sin\left(\frac{\pi}{4}\right)\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\sin x+\frac{\sqrt2}{2}\cdot cosx-\frac{\sqrt2}{2}\cdot\sin x+\frac{\sqrt2}{2}\cdot cosx=\frac{\sqrt2}{2}\cdot2\cdot cosx=\sqrt2\cdot cosx\)

a/ \(\frac{A}{2}+\left(\frac{B}{2}+\frac{C}{2}\right)=90^0\)

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

b/ \(\frac{tan^2A-tan^2B}{1-tan^2A.tan^2B}=\frac{\left(tanA-tanB\right)}{\left(1+tanA.tanB\right)}.\frac{\left(tanA+tanB\right)}{\left(1-tanA.tanB\right)}=tan\left(A-B\right).tan\left(A+B\right)\)

\(=tan\left(A-B\right).tan\left(180^0-C\right)=-tan\left(A-B\right).tanC\)

c/

\(A+B+C=180^0\Rightarrow cot\left(A+B\right)=-cotC\)

\(\Leftrightarrow\frac{cotA.cotB-1}{cotA+cotB}=-cotC\)

\(\Leftrightarrow cotA.cotB-1=-cotA.cotC-cotB.cotC\)

\(\Leftrightarrow cotA.cotB+cotB.cotC+cotA.cotC=1\)

Tại sao lại suy đc Tại sao lại suy đc c

phần chứng minh biểu thức không phụ thuộc \(x\)

ta có : \(A=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{sinacosa}{cota}=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{cos^2a}{cot^2a}\)

\(=\dfrac{cot^2a-cos^2a+cos^2a}{cot^2a}=\dfrac{cot^2a}{cot^2a}=1\left(đpcm\right)\)

ý còn lại : xem lại đề nha bn

phần chứng minh đẳng thức

ta có : \(\dfrac{sin2a-2sina}{sin2a+2sina}+tan^2\dfrac{a}{2}=\dfrac{2sinacosa-2sina}{2sinacosa+2sina}+tan^2\dfrac{a}{2}\)

\(=\dfrac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}+tan^2\dfrac{a}{2}=\dfrac{cosa-1}{cosa+1}+tan^2\dfrac{a}{2}\)

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

\(=-tan^2\dfrac{a}{2}+tan^2\dfrac{a}{2}=0\left(đpcm\right)\)

ta có : \(\dfrac{sina}{1+cosa}+\dfrac{1+cosa}{sina}=\dfrac{sin^2a+\left(1+cosa\right)^2}{sina\left(1+cosa\right)}\)

\(=\dfrac{sin^2a+cos^2a+2cosa+1}{sina\left(1+cosa\right)}=\dfrac{2cosa+2}{sina\left(cosa+1\right)}\)

\(=\dfrac{2\left(cosa+1\right)}{sina\left(cosa+1\right)}=\dfrac{2}{sina}\left(đpcm\right)\)

còn 2 câu kia để chừng nào rảnh mk giải cho nha

mk lm 2 câu còn lại nha

ta có : \(\dfrac{sin^2x}{sinx-cosx}-\dfrac{sinx+cosx}{tan^2x-1}=\dfrac{\left(1-cos^2x\right)\left(tan^2x-1\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)

\(=\dfrac{tan^2x-sin^2x-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\dfrac{sin^4x}{cos^2x}-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2-1\right)}\)

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

\(=sinx+cosx\left(đpcm\right)\)

ta có : \(\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-tan^2a.cot^2b}=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-\dfrac{sin^2a.cos^2b}{cos^2a.sin^2b}}\)

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

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

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

mk lm hơi tắc ! do tối rồi , mà mk lại đang ở quán nek nên không tiện làm dài . bạn thông cảm