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\(K=\frac{2sin\left(\frac{a+b}{2}\right).cos\left(\frac{a+b}{2}\right)+2sin\left(\frac{a+b}{2}\right).cos\left(\frac{a-b}{2}\right)}{2cos^2\left(\frac{a+b}{2}\right)-1+2cos\left(\frac{a+b}{2}\right).cos\left(\frac{a-b}{2}\right)+1}\)
\(K=\frac{sin\left(\frac{a+b}{2}\right)\left[cos\left(\frac{a+b}{2}\right)+cos\left(\frac{a-b}{2}\right)\right]}{cos\left(\frac{a+b}{2}\right)\left[cos\left(\frac{a+b}{2}\right)+cos\left(\frac{a-b}{2}\right)\right]}\)
\(K=\frac{sin\left(\frac{a+b}{2}\right)}{cos\left(\frac{a+b}{2}\right)}=tan\left(\frac{a+b}{2}\right)\)
a) \(\dfrac{tan\alpha-tan\beta}{cot\beta-cot\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}\)
\(=\dfrac{\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}}{\dfrac{cos\beta sin\alpha-cos\alpha sin\beta}{sin\beta sin\alpha}}\)
\(=\dfrac{sin\beta sin\alpha}{cos\beta cos\alpha}=tan\alpha tan\beta\).
b) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .
\(5sin\left(a+b-b\right)=3sin\left(a+b+b\right)\)
\(\Leftrightarrow5sin\left(a+b\right)cosb-5cos\left(a+b\right)sinb=3sin\left(a+b\right)cosb+3cos\left(a+b\right)sinb\)
\(\Leftrightarrow2sin\left(a+b\right)cosb=8cos\left(a+b\right)sinb\)
\(\Rightarrow\frac{sin\left(a+b\right)}{cos\left(a+b\right)}=\frac{4sinb}{cosb}\Rightarrow tan\left(a+b\right)=4tanb\)
2.
\(2sin\frac{A}{2}cos\frac{A}{2}=\frac{2sin\frac{B+C}{2}cos\frac{B-C}{2}}{2cos\frac{B+C}{2}cos\frac{B-C}{2}}=\frac{cos\frac{A}{2}}{sin\frac{A}{2}}\)
\(\Leftrightarrow2sin^2\frac{A}{2}=1\Leftrightarrow1-2sin^2\frac{A}{2}=0\)
\(\Leftrightarrow cosA=0\Rightarrow A=90^0\)
\(sina+sinb=2sin\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{2}}{2}\)
\(\Rightarrow sin\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{2}}{4}\) (1)
\(cosa+cosb=2cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{6}}{2}\)
\(\Rightarrow cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{6}}{4}\) (2)
(1); (2) \(\Rightarrow tan\left(\frac{a+b}{2}\right)=\frac{\sqrt{3}}{3}\) \(\Rightarrow tan\left(a+b\right)=\sqrt{3}\) \(\Rightarrow a+b=60^0\)
\(\Rightarrow sin\left(a+b\right)=sin\left(60^0\right)=\frac{\sqrt{3}}{2}\)
\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)
\(=2+2\left(cosa.cosb+sina.sinb\right)\)
\(=2+2.cos\left(a-b\right)=2+2.cos\frac{\pi}{3}=3\)
\(B=cos^2a+sin^2b+2cosa.sinb+cos^2b+sin^2a-2sina.cosb\)
\(=2-2\left(sina.cosb-cosa.sinb\right)\)
\(=2-2sin\left(a-b\right)=2-2sin\frac{\pi}{3}=2-\sqrt{3}\)