FORMULA FOR TAN A PLUS B

Problem 1 :

tan5𝜋12 - tan𝜋121 + tan5𝜋12 tan𝜋12

Solution :

Given, tan5𝜋12 - tan𝜋121 + tan5𝜋12 tan𝜋12tan (x - y) = tan x - tan y1 + tan x tan y= tan 5𝜋12 - 𝜋12= tan 4𝜋12= tan 𝜋3tan5𝜋12 - tan𝜋121 + tan5𝜋12 tan𝜋12= 3

Problem 2 :

tan13𝜋12 + tan7𝜋121 - tan13𝜋12 tan7𝜋12

Solution :

Given, tan13𝜋12 + tan7𝜋121 - tan13𝜋12 tan7𝜋12tan (x + y) = tan x + tan y1 - tan x tan y= tan 13𝜋12 + 7𝜋12= tan 20𝜋12= tan 5𝜋3= tan 2𝜋3 + 𝜋= tan 2𝜋3= -3tan13𝜋12 + tan7𝜋121 - tan13𝜋12 tan7𝜋12= -3

Problem 3 :

tan(a - b) + tan(b)1 - tan(a - b) tan(b)

Solution :

Given, tan(a - b) + tan(b)1 - tan(a - b) tan(b)tan (x + y) = tan x + tan y1 - tan x tan y= tan ((a - b) + b)= tan atan(a - b) + tan(b)1 - tan(a - b) tan(b)= tan a

Problem 4 :

tan(2c + 3d) - tan(d - c)1 + tan(2c + 3d) tan(d - c)

Solution :

Given, tan(2c + 3d) - tan(d - c)1 + tan(2c + 3d) tan(d - c)tan (x - y) = tan x - tan y1 + tan x tan y= tan ((2c + 3d) - (d - c))= tan (2c + 3d - d + c)= tan (3c + 2d)tan(2c + 3d) - tan(d - c)1 + tan(2c + 3d) tan(d - c)= tan (3c + 2d)

Problem 5 :

Given that tan (𝛼) = -4, evaluate tan 𝛼 + 3𝜋4.

Solution :

Given that tan (𝛼) = -4 tan 𝛼 + 3𝜋4 = tan 𝛼 + tan 3𝜋41 - tan (𝛼) tan 3𝜋4= -4 + (-1)1 - (-4) (-1)= -4 - 11- 4= -5-3= 53 tan 𝛼 + 3𝜋4 = 53

Problem 6 :

Given that tan (𝛽) = 2, evaluate tan 𝜋4 - 𝛽.

Solution :

Given that tan (𝛽) = 2 tan 𝜋4 - 𝛽 = tan 𝜋4 - tan 𝛽1 + tan 𝜋4 tan 𝛽= 1 - 21 + (1) (2)= - 1`1+ 2= -13= -1 tan 𝜋4 - 𝛽 = -1

Problem 7 :

Given that tan (x) = 23, evaluate tan x - 5𝜋4.

Solution :

Given that tan (x) = 23 tan x - 5𝜋4 = tan x - tan 5𝜋41 + tan x tan 5𝜋4= 23 - 11 + 23 (1)= 2 - 331+ 23= -1353= -13 × 35= -15 tan x - 5𝜋4 = -15

Problem 8 :

Given that tan (x) = -15, evaluate tan x + 7𝜋4.

Solution  :

Given that tan (x) = -15 tan x + 7𝜋4= tan x + tan 7𝜋41 - tan x tan 7𝜋4= -15 + (-1)1 - -15 (-1)= -15 - 11- 15= -1 - 555 - 15= -6545= -65 × 54= -64 tan x + 7𝜋4 = -64

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