Prove that :cot 15/2 degrees = square root of 2 + root of 3 + root of 4 + root of 6
Answers
Answered by
13
cotx=1+cosx/sinx
Cot15=1+cos30/sin30=(1+√3/2)/1/2=2+√3
cot15/2=cot15+√1+cot^2(15)=2+√3+√1+(2+√3)^2
=2+√3+√1+7+4√3=2+√3+√8+4√3
2+√3+√2(√4+2√3)=√4+√3+√2√(√3+1)^2 (4+2√3=(√3+1)^2 )
=√3+√4+√2(√3+1)=√3+√4+√6+√2
=√2+√3+√4+√6
Cot15=1+cos30/sin30=(1+√3/2)/1/2=2+√3
cot15/2=cot15+√1+cot^2(15)=2+√3+√1+(2+√3)^2
=2+√3+√1+7+4√3=2+√3+√8+4√3
2+√3+√2(√4+2√3)=√4+√3+√2√(√3+1)^2 (4+2√3=(√3+1)^2 )
=√3+√4+√2(√3+1)=√3+√4+√6+√2
=√2+√3+√4+√6
Similar questions