Math, asked by ashu5988, 1 year ago

tanx+tan(60+x)-tan(60-x)​

Answers

Answered by agarwalkritika004555
7

LHS = tan(x)*tan(60-x)*tan(60+x)=tan(x)[(tan(60)-tan(x))/(1+tan(60*tan(x))]*[(tan(60)+tan(x))/(1-tan(60*tan(x))]=tan(x)[√3-tan(x)/(1+√3*tan(x))]*[√3+tan(x)/(1-√3*tan(x))] =tan(x)*[(3-tan²(x) )/(1–3tan²(x))] = [(3tan(x)-tan³(x) )/(1–3tan²(x))] = tan(3x).

In other words, tan(3x) = 1/√3 = tan(30°) or x=10°

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