Question

HS ( =frac{1}{tan 3 alpha-tan alpha} )
( frac{1}{frac{1}{cot 3 alpha}-frac{1}{cot alpha}}-frac{1}{cot 3 alpha-cot alpha} )
1
( =frac{cot alpha cot 3 alpha}{cot alpha-cot 3 alpha}+frac{1}{cot alpha-cot 3 alpha} )
( =frac{1+cot alpha cot 3 alpha}{cot alpha-cot 3 alpha} )
( frac{1+cos alpha cos 3 alpha}{frac{sin alpha sin 3 alpha}{cos alpha}-frac{cos 3 alpha}{sin alpha}} )
( =cos alpha cos 3 alpha+sin alpha sin 3 alpha )
( sin 3 alpha cos alpha-cos 3 alpha sin alpha )
( frac{cos (3 alpha-alpha)}{sin (3 alpha-alpha)}=frac{cos 2 alpha}{sin 2 alpha} )
formulae cos ( A ) ass ( +sin A sin B=cos (A-B) quad=cot 2 alpha )
( sin A cos B-cos A sin B=sin (A-R) )

# _ Prove that to atan 3a - tana 1 - = cot 2a. cot3a-cota

Solution