Please help with math trigonometry?

If cos a = -sqrt10 / 10 and a is in the third quadrant. Calculate tan (3π/2 - a)

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  • 1 month ago

    Well,

    sin a = sqrt( 1 - cos^2(a))  = sqrt( 1 - 10/100) = sqrt( 90/100) = -sqrt90 /10

    tan a = sin a/cos a = sqrt(9) = 3

    tan(b-a) = (tan b - tan a)/(1 + tan a tan b), but unfortunately tan(3pi/2) is infinite.

    sin(b-a) = sin b cos a - cos b sin a

    if b = 3pi/2, sin b = -1, cos b = 0, so sin(3pi/2 - a) = -cos a = sqrt10/10

    cos(b-a) = cos b cos a + sin b sin a

    cos(3pi/2 - a) = -sin a = sqrt(90)/10

    so tan(3pi/2 -a) = (sqrt10/10)/(sqrt(90)/10) = 1/3

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