Tan75/2 ka solution dena sir lekin vistar se

Answer 1

Is that Hindi?

Fortunately mathematics is a universal language.

from tan 2A = 2tanA/(1 - tan^2 A)
tan 75 = 2tan (75/2) /(1 - tan^2 (75/2) )

we know tan 75 = sin 75/cos 75
= sin(45+30)/cos(45+30)
= sin45cos30 + cos45sin30)/(cos45cos30 - sin45sin30)
= ( (√2/2)(√3/2) + (√2/2)(1/2))/( (√2/2)(√3/2) - (√2/2)(1/2) )
= (√6/4 + √2/4)/(√6/4 - √2/4)
= (√6 + √2)/(√6 - √2)

let x = tan 75/2°
then (√6 + √2)/(√6 - √2) = 2x/(1 - x^2)
2x(√6-√2) = √6 + √2 - (√6 + √2) x^2

(√6+√2) x^2 + 2(√6-√2) x - (√6+√2) = 0

x = ( -2(√6-√2) ± √(4(√6-√2)^2 + 4(√6+√2)(√6+√2) )/(2(√6-√2) )
= (-2(√6-√2) ± 8) / (2(√6+√2))
= ( -√6 + √2 ± 8)/(√6 + √2)

but since 75/2 is in the first quadrant, the answer must be positive, so

tan (75/2)° = (-√6 + √2 + 8)/(√6 + √2)

argghhhh!!!!
I made an error somewhere, I should have written it out in full first. Perhaps you can see where I went wrong.
I know my formula is correct and tan 75 is correct as (√6 + √2)/(√6 - √2)

Answer 2

Using (√6+√2) x^2 + 2(√6-√2) x - (√6+√2) = 0
I used Wolfram to solve and it gave me the correct answer of 0.76733
so my error must be in the use of the quadratic formula.

http://www.wolframalpha.com/input/?i=solve+(%E2%88%9A6%2B%E2%88%9A2)+x%5E2+%2B+2(%E2%88%9A6-%E2%88%9A2)+x+-+(%E2%88%9A6%2B%E2%88%9A2)+%3D+0