tanA=kcotB then cos(A+B)/cos(A-B)
1.1+k/1-k 2.1-k/1+k 3.k+1/k-1 4.k-1/k+1
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trig: solving tanA=kcotB then cos(A+B)/cos(A-B) 
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trig: solving tanA=kcotB then cos(A+B)/cos(A-B)
by yamini on Tue Aug 11, 2009 3:05 pm
can u please help me solving this problem
tanA=kcotB then cos(A+B)/cos(A-B)
1.1+k/1-k 2.1-k/1+k 3.k+1/k-1 4.k-1/k+1
tanA=kcotB then cos(A+B)/cos(A-B)
1.1+k/1-k 2.1-k/1+k 3.k+1/k-1 4.k-1/k+1
- yamini
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- Joined: Mon Aug 10, 2009 1:25 pm
by stapel_eliz on Tue Aug 11, 2009 6:14 pm
A good start might be to notice that cotangent is the reciprocal of the tangent:
. . . . .\, =\, \frac{1}{\tan(B)})
See if this helps with the application of angle-sum identities....
. . . . .
See if this helps with the application of angle-sum identities....
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stapel_eliz - Posts: 1703
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