Question

tan 15 degree is equilvalent to​

Answer 1

$\mathfrak{ANSWER}\\\tan 15^{\circ} = \tan (45-30)^{\circ}\\By\: the\: trigonometry\: formula\: , \: we \: know,\\\tan(A-B) = \frac{(tanA-tanB)}{(1=tanA.tanB)}\\\therefore \: ,WE\: CAN\:WRITE\\tan(45-30)^{\circ} = \frac{tan45^{\circ}-tan30^{\circ}}{1+tan45^{\circ}.tan30^{\circ}}\\Now \: putting\: the\: values \: of\: tan 45^{\circ} \: and \: tan 30^{\circ} \: from\: the \: table\: we \: get;\\tan(45-30)^{\circ} = \frac{(1-\frac{1}{\sqrt{3}})}{(1+1.\frac{1}{\sqrt{3}})}$$\\tan(15^{\circ}) = \frac{\sqrt{3}-1}{\sqrt{3}+1}\\HERE \:,\:THE \:VALUE \:OF\:tan(15^{\circ})\:IS\:\frac{\sqrt{3}-1}{\sqrt{3}+1}\\\\BY\:PUTTING\:THE\:VALUE\:OF\:\sqrt{3}\,\:WHICH\:IS\:EQUAL\:TO\:1.732\\\therefore \:tan(15^{\circ})= \frac{1.732-1}{1.732+1}\\tan(15^{\circ})= 0.2679\\APPROXIMATELY \:tan(15^{\circ})\approx 0.27\\\\\textbf{HOPE\:IT\:MAY\:HELP\:YOU...}$