Question

prove that?
tan10-tan50+tan70= (3^)1/2

Answer 1

tan70 = tan(60+10) use the identity : tan(a+b) = (tan(a) + tan(b))/(1-tan(a)tan(b)) so, tan70 = (tan60+tan10)/(1-tan60tan10) tan70 = (sqrt(3)+tan(10))/(1-sqrt(3)tan10) ... (1) tan50 = tan(60-10) use the identity : tan(a-b) = (tan(a) - tan(b))/(1+tan(a)tan(b)) so, tan50 = (tan60-tan10)/(1+tan60tan10) tan50 = (sqrt(3)-tan(10))/(1+sqrt(3)tan10) ... (2) therefore, the value of tan70 - tan50 = (sqrt(3)+tan(10))/(1-sqrt(3)tan10) - (sqrt(3)+tan(10))/(1-sqrt(3)tan10) if that simplied becomes, we get tan70 - tan50 = 8tan10/(1-3tan^2 10)

now, look at the original question above is , sin10-sin50+tan70 which similar with tan70 - tan50 + tan10. (the value of tan70-tan50 already done) so, tan70 - tan50 + tan10 = 8tan10/(1-3tan^2 10) + tan10 = (8tan10+tan10-3tan^3 10)/(1-3tan^2 10) = (9tan10-3tan^3 10)/(1-3tan^2 10) = 3 (3tan10 - tan^3 10)/(1-3tan^2 10)

then, remember that there is a formula in trigono : tan(3x) = (3tan(x) - tan^3 (x))/(1 - 3tan^2 (x))


therefore, the calculation above can be simplied becomes 3 (3tan10 - tan^3 10)/(1-3tan^2 10) = 3 tan(3*10) = 3 tan30 = 3 * 1/3 sqrt(3) = sqrt(3)

Answer 2

Answer: root 3

Step-by-step explanation:

Formulae used :

1)tan (A+B)= (tan A + tan B)/(1-tanAtanB)

2)tan (A - B)=(tan A - tan B)/(1+tanAtanB)

3)tan 3x= (3tanx - tan^3 x)/(1-3tan^2 x)

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