# 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

• 64
tan 50 = 2 tan 10 + tan40
• -7
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• -25
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
• -11

• 27
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
• 10
TAN 50=TAN(40+10)
THEN APPLY TAN(A+B) IDENTITY
• 4
this is the easiest way to solve this question

• 2
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