Posted by Dee on Monday, November 8, 2010 at 11:40pm.
Find the exact value of tan(ab)
sin a = 4/5, 3pi/2<a<pi;
tan b = sqrt2, pi/2<b<pi
identity used is:
tan(ab)=(tan atan b)/1+tan a tan b
simplify answer using radicals.
(a is alpha, b is beta)

Trigonometry  Reiny, Monday, November 8, 2010 at 11:54pm
Both the sine and cosine curves are the same for
3pi/2<a<pi as they are for π/2<a<π (2nd quad)
so if sina = 4/5, then cosa = 3/5
and tana = 4/3
then tan(ab)
= (tana  tanb)/(1 + tanatanb)
= ((4/3)  (√2))/( 1 + (4/3)(√2))
= (√2  4/3)/(1 + 4√2/3)
= (3√2  4)/(3 + 4√2)
I don't know if you have to rationalize that, if you do carefully multiply top and bottom by (3  4√2)

Trigonometry  Dee, Tuesday, November 9, 2010 at 12:16am
I put that in but it said the answer was
3625ã2/23
any idea how that works?

Trigonometry  Reiny, Tuesday, November 9, 2010 at 12:40am
That is exactly what my answer works out to if you rationalize it.
I had suggested to do that.

Trigonometry  Anonymous, Thursday, October 20, 2016 at 8:10pm
8 cot(A) − 8/
1 + tan(−A)
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