Verify if tan(a+b)-tana*tanb*tan(a+b)-tan a- tanb=0?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to verify tan(a+b) - tan a*tan b*tan(a+b) - tan a - tan b =0

tan(a+b) - tan a*tan b*tan(a+b) - tan a - tan b

=> tan(a + b)[1 - tan a* tan b] - tan a - tan b

use tan (a + b) = (tan a + tan b)/(1 - tan a* tan b)

=>(tan a + tan b)/(1 - tan a* tan b)*(1 - tan a*tan b) - tan a - tan b

=>(tan a + tan b) - tan a - tan b

=>(tan a - tan a + tan b - tan b

=> 0

This proves that tan(a+b) - tan a*tan b*tan(a+b) - tan a - tan b = 0

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll factorize by tan(a+b):

tan(a+b)(1 - tan a*tan b) - tan a - tan b = 0

We'll move tan a and tan b to the right side:

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

We'll divide by (1 - tan a*tan b):

tan(a+b) = (tan a + tan b)/(1 - tan a*tan b)

But the function tangent tan (a+b) is the ratio:

tan(a+b) = sin(a+b)/cos(a+b)

We'll write the formulas for the sine and cosine of the sum of angles a and b:

sin(a+b) = sina*cosb + sinb*cosa

cos(a+b) = cosa*cosb - sina*sinb

We'll substitute sin(a+b) and cos(a+b) by their formulas:

tan(a+b) = (sina*cosb + sinb*cosa)/(cosa*cosb - sina*sinb)

We'll factorize by cosa*cosb:

tan(a+b) =cosa*cosb*[(sina*cosb/cosa*cosb) + (sinb*cosa/cosa*cosb)]/cosa*cosb*[1 - (sina*sinb/cosa*cosb)]

We'll simplify and we'll get:

tan(a+b) = (sina/cos a + sinb/cos b)/(1 - tan a*tan b)

tan(a+b) = (tan a + tan b)/(1 - tan a*tan b)

The identity tan(a+b)(1 - tan a*tan b) - tan a - tan b = 0 is verified.

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