(7) R.H.S = tan(60° + A) tan(60°-A)

= \(\cfrac{tan\,60°+tan\,A}{1-tan\,60°\,tan\,A}\) x \(\cfrac{tan\,60°-tan\,A}{1+tan\,60°\,tan\,A}\)

= \(\cfrac{\sqrt3+tan\,A}{1-\sqrt3\,tan\,A}\) x \(\cfrac{\sqrt3-tan\,A}{1+\sqrt3\,tan\,A}\)

= \(\cfrac{(\sqrt3)^2+tan^2\,A}{1-(\sqrt3\,tan\,A)^2}\) (\(\because\) (a + b)(a - b) = a² - b²)

= \(\cfrac{3-tan^2A}{1-3\,tan^2A}\)

= \(\cfrac{\cfrac{3-sin^2A}{cos^2A}}{\cfrac{1-3\,sin^2A}{cos^2A}}\)

= \(\cfrac{3\,cos^2A-sin^2A}{cos^2A-3\,sin^2A}\) 

= \(\cfrac{2\,cos^2A+1-sin^2A-sin^2A}{cos^2A-2sin^2A-(1-cos^2A)}\)

= \(\cfrac{2(cos^2A-sin^2A)+1}{2(cos^2A-sin^2A)-1}\)

= \(\cfrac{2\,cos^2A+1}{2\,cos^2A-1}\)

= L.H.S

Hence proved

(8) cot x cot 2x - cot 2x cot 3x - cot 3x cot x

= cot x cot 2x - cot 3x (cot 2x + cot x)

= cot x cot 2x - cot (2x + x) (cot 2x + cot x)

= cot x cot 2x - \(\left(\cfrac{cot \,2x\,cot \,x-1}{cot\,x+cot\,2x}\right)\)(cot x + cot 2x)

(\(\because\) cot (A - B))

= \(\cfrac{cot\,A\,cotB-1}{cot\,B+cot\,A}\)

= cot x cot 2x - (cot x cot 2x - 1)

= cot x cot 2x - cot x cot 2x + 1

= 1

Hence proved