Question

(i)Show that cot 75 degree = 2- sqrt 3
(ii) Hence, or otherwise show that cosec^2 75 degree = 4 cot 75 degree

Answer 1

firstly, cot=cos/sin. and after all, you can use the trigo formula.
split 75=45+30
so cos(30+45)=cos45cos30-sin45sin30
sin(30+45)=sin30cos45+sin45cos30

Answer 2

Thanks. how to solve for part ii?

Answer 3

cos(30+45)=cos45cos30-sin45sin30=sqrt2/2*sqrt3/2-sqrt2/2*1/2
sin(30+45)=sin30cos45+sin45cos30=1/2*sqrt2/2+sqrt2/2*sqrt3/2

Answer 4

so you need to use, cos(30+45)/sin(30+45)=(sqrt2+sqrt6)/(sqrt6-sqrt2) (simplified, both numerator and denominator times 4)
so now, you can times (sqrt2 + sqrt6) for both numerator and denominator,
it will give you (sqrt2 + sqrt6)^2/4 which is 2+ sqrt3

Answer 5

i thought is minus?

Answer 6

oh, yest. just a typo.
so you need to use, cos(30+45)/sin(30+45)=(sqrt6-sqrt2)/(sqrt6+sqrt2)
so now, you can times (sqrt2 - sqrt6) for both numerator and denominator,
it will give you (sqrt2 - sqrt6)^2/4 which is 2+ sqrt3

Answer 7

which gives you 2- sqrt3

Answer 8

for the second part, we all know that cosec^2=1 + cot^2.
So just sub in the formula,
4cot75= 4(2-sqrt3)=8-4sqrt3
1+cot^2=1 + (2-sqrt3)^2=1+4+3-4sqrt3=8-4sqrt3

Answer 9

for the second part, we all know that cosec^2=1 + cot^2.
So just sub in the formula,
4cot75= 4(2-sqrt3)=8-4sqrt3
1+cot75^2=1 + (2-sqrt3)^2=1+4+3-4sqrt3=8-4sqrt3

Answer 10

Thanks fireofsky.