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prove that tan 50 = 2 ta n 10 + ta n 40

tan50= tan(40+10)

we have tan(a+b) = (tan a +tan b)/(1-tan a tan b)

= tan(40+10) = (tan40+ tan10)/(1-tan40tan10)

= tan50 = (tan40+ tan10)/(1-tan40tan10)

= tan50(1-tan40tan10)=tan40+ tan10

= tan50-tan50tan40tan10 = tan40+tan10

and tan50 = tan(90-40) = cot40

= tan50-cot40tan40tan10 = tan40+ tan10

= tan50-tan10 = tan40+ tan10

because cot40= 1/tan40

= tan50 = tan40+2tan10

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tan 50 = 2 tan 10 + tan40
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Use the identities: 
(1) tan(90-x)=1/tanx 
(2) tan(2x)=2tanx/(1-tan^2(x)) 

Using (2) with x=40: 
tan80 = 2tan40/(1-tan^2(40)) 
(tan80)/2 = tan40/(1-tan^2(40)) 
2/tan80 = (1-tan^2(40))/tan40 
2/tan80 = 1/tan40 - tan40 
1/tan40 = tan40 + 2/tan80 

Using (1), LHS=tan50 & 1/tan80=tan10 
And this gives us: 
tan50 = tan40 +2tan10
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Please find this answer

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we have tan(a+b) = (tan a +tan b)/(1-tan a tan b) 

tan(40+10) = (tan40+ tan10)/(1-tan40tan10) 

tan50 = (tan40+ tan10)/(1-tan40tan10) 

tan50(1-tan40tan10)=tan40+ tan10 

tan50-tan50tan40tan10 = tan40+tan10 

and tan50 = tan(90-40) = cot40 

=> tan50-cot40tan40tan10 = tan40+ tan10 

tan50-tan10 = tan40+ tan10 

because cot40= 1/tan40 

tan50 = tan40+2tan10
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TAN 50=TAN(40+10)
THEN APPLY TAN(A+B) IDENTITY
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this is the easiest way to solve this question

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