1+tan30/1-tan30=2+tan60
Answers
Answered by
1
tan 30 = 1/√3 and tan 60 = √3
Substituting the values
(1 + 1/√3)/(1-1/√3) = 2 + √3
[(√3+1)/√3] / [(√3-1)/√3)] = 2+√3
√3+1/√3-1 = 2 + √3
On rationalizing
√3+1/√3-1 * √3+1/√3+1 = 2+√3
(√3+1)²/(√3)² - 1 = 2+√3
3 + 1 + 2√3/3-1 = 2+√3
4+2√3 /2 = 2+√3
2+√3 = 2+√3
Hence proved
Substituting the values
(1 + 1/√3)/(1-1/√3) = 2 + √3
[(√3+1)/√3] / [(√3-1)/√3)] = 2+√3
√3+1/√3-1 = 2 + √3
On rationalizing
√3+1/√3-1 * √3+1/√3+1 = 2+√3
(√3+1)²/(√3)² - 1 = 2+√3
3 + 1 + 2√3/3-1 = 2+√3
4+2√3 /2 = 2+√3
2+√3 = 2+√3
Hence proved
kartik270:
i have't understand the second last solution can u make me understand plz
it is rationalising
Similar questions