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29 July, 22:51

If sin a = 10 / 10 with a in qi and tan b = 4 3 with b in qi, find tan (a +

b. and cot (a + b).

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Answers (1)
  1. 30 July, 01:17
    0
    Tan (A+B) = [tan (A) + tan (B) ]/[1-tan (A) * tan (B) ]

    also

    cot (A+B) = 1/[cot (A+B) ]

    but

    sin (A) = √10/10=y/r

    so

    y=√10 and r=10

    to find x we use the Pythagorean theorem:

    x^2=r^2+y^2

    x^2=10^2 - (√10) ^2

    x^2=100-10

    x^2=90

    x=√90

    x=3√10

    since:

    tan (A) = y/x=√10 / (3√10) = 1/3

    also

    tan B is 4/3 hence:

    cot (A+B) = 1/[cot (A+B) ]

    but

    tan (a+b) = [tan (a) + tan (b) ]/[1-tan (a) tan (b) ]

    =[1/3+4/3]/[1 - (1/3) (4/3) ]

    = (5/3) / (1-4/9)

    = (5/3) / (5/9)

    =15/5

    =3

    but

    cot x=1/tan x

    so

    cot (A+B) will be 1/3
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