L.H.S.\(\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}\)
Using Cosine formula, we get
\(\frac{\frac{b^2+c^2-a^2}{2bc}}{a}+\frac{\frac{a^2+c^2-b^2}{2ac}}{b}+\frac{\frac{a^2+b^2-c^2}{2ab}}{c}\)
\(\Rightarrow\frac{b^2+c^2-a^2}{2abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}\)
\(=\frac{1}{2abc}[b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2]\)
\(\Rightarrow\frac{a^2+b^2+c^2}{2abc}R.H.S.\)

Using Sine formula
\(\frac{a}{sin A}=\frac{b}{sinB}=\frac{c}{sinC}=k(say)\)
R.H.S.\(\frac{b-c}{2}.cos\frac{A}{2}\)
\(=\frac{ksinB-ksinC}{ksinA}.cos\frac{A}{2}\)
\(=\frac{sinB-sinC}{sinA}.cos\frac{A}{2}\)
\(=\frac{2cos(\frac{B+C}{2})sin(\frac{B-C}{2})}{2sin\frac{A}{2}.cos\frac{A}{2}}.cos\frac{A}{2}\)
\(=\frac{cos(\frac{\pi}{2}-\frac{A}{2}).sin(\frac{B-C}{2})}{sin\frac{A}{2}}\)
\(\Rightarrow\frac{sin\frac{A}{2}.sin(\frac{B-C}{2})}{sin\frac{A}{2}}\)
\(\Rightarrow sin(\frac{B-C}{2})L.H.S.\)

\({-2\over3}\)

Here cot x = \(3\over4\) tan x =\({1\over cot x}={4\over 3}\)
Now sec²x = 1 + tan² x
\(\Rightarrow sec^2x=1+({4\over3})^2\)
\(\Rightarrow sec^2x=1+{16\over 9}\)
\(\Rightarrow sec^2 x={25\over 9}\Rightarrow sec \ x= \pm {5\over3}\)
But x lies in third quadrant.
\(\therefore sec \ x ={-5\over 3}\)
\(cos x={1\over sec x}={-3\over5}\)
Also sin² x + cos²x= 1
\(\Rightarrow sin^2x+({-3\over 5})^2=1\)
\(\Rightarrow sin^2x=1-{9\over 25}\)
\(\Rightarrow sin^2x={16\over25}\Rightarrow sin x=\pm{4\over5}\)
But x lies in third quadrant.
\(\therefore sin \ x={-4\over 5}\)
\(\therefore cosec \ x={1\over sin \ x}={-5\over 4}\)

Here cos x = \(-{1\over 2}\)
Now sin² x + cos²x = 1
\(\Rightarrow sin^2+(-{1\over 2})^2=1\)
\(\Rightarrow sin^2x=1-{1\over 4}\)
\(\Rightarrow sin^2x={3\over4}\)
\(\Rightarrow sinx=\pm{\sqrt{3}\over2}\)
But x lies in third quadrant.
\(\therefore sin x=-{\sqrt{3}\over2}\)
Now \(cosec \ x={1\over sin \ x}=-{2\over \sqrt{3}}\)
and \(sec \ x ={1\over cos \ x}=-2\)
\(tan \ x={sin \ x\over cos \ x}={-\sqrt{3/2}\over-1/2}=\sqrt{3}\)
\(and \ cot x={cos \ x \over sin \ x}={-1/2\over -\sqrt{3}/2}={1\over \sqrt{3}}\)

\({2\pi\over9},{\pi\over3},{4\pi\over9}\)

We have
L.H.S. = \({({sin \ 7x}+ sin 5x) +(sin 9x + sin 3x)\over (cos 7x + cos 5x) + (cos 9x + cos 3x)}\)
\(={[2sin({7x+5x\over2})cos({7x-5x\over2})]+[2sin({9x+3x\over2})cos({9x-3x\over2})]\over[2cos({7x+5x\over2})cos({7x-5x\over2})]+[2cos({9x+3x\over2})cos({9x-3x\over2})]}\)
\(={{2sin \ 6x}cosx +2sin \ 6xcos3x \over 2cos \ 6x cosx +2cos6x cos 3x}\)
\(={{2sin \ 6x}(cosx +cos3x) \over 2cos \ 6x (cosx +cos3x )}\)
\(={sin6x\over cos6x}=\tan6x=R.H.S\).

sin x + sin 3x + sin 5x = 0
\(\Rightarrow\)(sin 5x + sin x) + sin 3x = 0
\(\Rightarrow\)2sin \(({5x+x\over 2})cos ({5x+x\over 2})+sin 3x=0\)
\(\Rightarrow\)2 sin 3x cos 2x + sin 3x = 0
\(\Rightarrow\)sin 3x (2 cos 2x + 1) = 0
\(\Rightarrow\)Either sin 3x = 0 or 2 cos 2x + 1 = 0
\(\Rightarrow\)3x = n\(\pi\) or cos 2x=\(-{1\over2}=cos {2\pi\over3},n\in Z\)
\(\Rightarrow\) \(x={n\pi\over3} or 2x=2n\pi\pm{2\pi\over3},n\in z.\)
\(\Rightarrow\) \(x={n\pi\over3} or x=n\pi\pm{\pi\over3},n\in z.\)

We have
L.H.S. = \({sin \ x + sin \ 3x\over cos \ x + cos \ 3x}\)
\(={2sin({x+3x\over 2})cos({x-3x\over 2})\over 2cos ({x+3x\over 2})cos({x-3x\over2})}\)
\(\left[\begin{array}{l} \because \sin \mathrm{C}+\sin \mathrm{D}=2 \sin \left(\frac{\mathrm{C}+\mathbf{D}}{2}\right) \cos \left(\frac{\mathbf{C}-\mathbf{D}}{2}\right) \\ \cos \mathrm{C}+\cos \mathrm{D}=2 \cos \left(\frac{\mathbf{C}+\mathrm{D}}{2}\right) \cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right) \end{array}\right]\)
\(={2 \ sin \ 2x \ cos \ (-x)\over 2 \ cos \ 2x \ cos (-x) }\)
= tan 2x = R.H.S.