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 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 \(sinx< 0\Leftrightarrow\pi< x< 2\pi\)
\(\Leftrightarrow B=\dfrac{1}{sinx}.\dfrac{\sqrt[]{2}}{-sinx}-\sqrt[]{2}\)
\(\Leftrightarrow B=-\dfrac{\sqrt[]{2}}{sin^2x}-\sqrt[]{2}\)
\(\Leftrightarrow B=-\sqrt[]{2}\left(\dfrac{1}{sin^2x}+1\right)\)
\(\Leftrightarrow B=-\sqrt[]{2}.\left(2+cot^2x\right)\)
