1.

tan 75°_cot75° = 4sin 600

Answer»

tan75°=tan(30°+45°)

=(tan30°+tan45°)/(1-tan30°tan45°)

={(1/√3)+1}/{1-(1/√3)(1)}

=(√3+1)/(√3-1)

So,cot75°=1/tan75°

=(√3-1)/(√3+1)

Then,tan75°+cot75°={(√3+1)2+(√3-1)2}/{(√3)2-(1)2}

=(3+1+2√3+3+1-2√3)/(3-1)

=4.


Discussion

No Comment Found

Related InterviewSolutions