Câu hỏi của Lê Đăng Hải Phong

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Nguyễn Đức Trí · Accepted Answer

Câu 2 :$\dfrac{sin^3\dfrac{B}{2}}{cos\left(\dfrac{A+C}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(\dfrac{A+C}{2}\right)}+\dfrac{cos\left(A+C\right)}{sinB}.tanB=2$ $VT=\dfrac{sin^3\dfrac{B}{2}}{cos\left(\dfrac{180^o-B}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(\dfrac{180^o-B}{2}\right)}-\dfrac{cos\left(180^o-B\right)}{sinB}.tanB$$\Leftrightarrow VT=\dfrac{sin^3\dfrac{B}{2}}{cos\left(90^o-\dfrac{B}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(90^o-\dfrac{B}{2}\right)}-\dfrac{\left(-cosB\right)}{sinB}.tanB$$\Leftrightarrow VT=\dfrac{sin^3\dfrac{B}{2}}{sin\dfrac{B}{2}}+\dfrac{cos^3\dfrac{B}{2}}{cos\dfrac{B}{2}}+\dfrac{sinB}{sinB}$$\Leftrightarrow VT=sin^2\dfrac{B}{2}+cos^2\dfrac{B}{2}+1$$\Leftrightarrow VT=1+1=2=VP$$\Rightarrow dpcm$Câu trả lời: Câu 2 :$\dfrac{sin^3\dfrac{B}{2}}{cos\left(\dfrac{A+C}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(\dfrac{A+C}{2}\right)}+\dfrac{cos\left(A+C\right)}{sinB}.tanB=2$ $VT=\dfrac{sin^3\dfrac{B}{2}}{cos\left(\dfrac{180^o-B}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(\dfrac{180^o-B}{2}\right)}-\dfrac{cos\left(180^o-B\right)}{sinB}.tanB$$\Leftrightarrow VT=\dfrac{sin^3\dfrac{B}{2}}{cos\left(90^o-\dfrac{B}{2}\right)}+\dfrac{cos^3\dfrac{B}{2}}{sin\left(90^o-\dfrac{B}{2}\right)}-\dfrac{\left(-cosB\right)}{sinB}.tanB$$\Leftrightarrow VT=\dfrac{sin^3\dfrac{B}{2}}{sin\dfrac{B}{2}}+\dfrac{cos^3\dfrac{B}{2}}{cos\dfrac{B}{2}}+\dfrac{sinB}{sinB}$$\Leftrightarrow VT=sin^2\dfrac{B}{2}+cos^2\dfrac{B}{2}+1$$\Leftrightarrow VT=1+1=2=VP$$\Rightarrow dpcm$

Nguyễn Đức Trí · Answer

Câu 3 :a) $A=sin\left(90^o-x\right)+cos\left(180^o-x\right)-sin^2x\left(1+tan^2x\right)-tan^2x$$\Leftrightarrow A=cosx-cosx+\dfrac{sin^2x}{cos^2x}-tan^2x$$\Leftrightarrow A=tan^2x-tan^2x=0$b) $B=\dfrac{1}{sinx}\sqrt[]{\dfrac{1}{1+cosx}+\dfrac{1}{1-cosx}}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{1}{sinx}\sqrt[]{\dfrac{1-cosx+1+cosx}{\left(1+cosx\right)\left(1-cosx\right)}}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{1}{sinx}\sqrt[]{\dfrac{2}{\left(1-cos^2x\right)}}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{1}{sinx}\sqrt[]{\dfrac{2}{sin^2x}}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{1}{sinx}.\dfrac{\sqrt[]{2}}{\left|sinx\right|}-\sqrt[]{2}$- Với $sinx>0\Leftrightarrow0< x< \pi$$\Leftrightarrow B=\dfrac{1}{sinx}.\dfrac{\sqrt[]{2}}{sinx}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{\sqrt[]{2}}{sin^2x}-\sqrt[]{2}$$\Leftrightarrow B=\sqrt[]{2}\left(\dfrac{1}{sin^2x}-1\right)$$\Leftrightarrow B=\sqrt[]{2}.cot^2x$- Với $sinx< 0\Leftrightarrow\pi< x< 2\pi$$\Leftrightarrow B=\dfrac{1}{sinx}.\dfrac{\sqrt[]{2}}{-sinx}-\sqrt[]{2}$$\Leftrightarrow B=-\dfrac{\sqrt[]{2}}{sin^2x}-\sqrt[]{2}$$\Leftrightarrow B=-\sqrt[]{2}\left(\dfrac{1}{sin^2x}+1\right)$$\Leftrightarrow B=-\sqrt[]{2}.\left(2+cot^2x\right)$Câu trả lời: Câu 3 :a) $A=sin\left(90^o-x\right)+cos\left(180^o-x\right)-sin^2x\left(1+tan^2x\right)-tan^2x$$\Leftrightarrow A=cosx-cosx+\dfrac{sin^2x}{cos^2x}-tan^2x$$\Leftrightarrow A=tan^2x-tan^2x=0$b) $B=\dfrac{1}{sinx}\sqrt[]{\dfrac{1}{1+cosx}+\dfrac{1}{1-cosx}}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{1}{sinx}\sqrt[]{\dfrac{1-cosx+1+cosx}{\left(1+cosx\right)\left(1-cosx\right)}}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{1}{sinx}\sqrt[]{\dfrac{2}{\left(1-cos^2x\right)}}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{1}{sinx}\sqrt[]{\dfrac{2}{sin^2x}}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{1}{sinx}.\dfrac{\sqrt[]{2}}{\left|sinx\right|}-\sqrt[]{2}$- Với $sinx>0\Leftrightarrow0< x< \pi$$\Leftrightarrow B=\dfrac{1}{sinx}.\dfrac{\sqrt[]{2}}{sinx}-\sqrt[]{2}$$\Leftrightarrow B=\dfrac{\sqrt[]{2}}{sin^2x}-\sqrt[]{2}$$\Leftrightarrow B=\sqrt[]{2}\left(\dfrac{1}{sin^2x}-1\right)$$\Leftrightarrow B=\sqrt[]{2}.cot^2x$- Với $sinx< 0\Leftrightarrow\pi< x< 2\pi$$\Leftrightarrow B=\dfrac{1}{sinx}.\dfrac{\sqrt[]{2}}{-sinx}-\sqrt[]{2}$$\Leftrightarrow B=-\dfrac{\sqrt[]{2}}{sin^2x}-\sqrt[]{2}$$\Leftrightarrow B=-\sqrt[]{2}\left(\dfrac{1}{sin^2x}+1\right)$$\Leftrightarrow B=-\sqrt[]{2}.\left(2+cot^2x\right)$

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