trignometric functions:solve

Asked by  | 7th Oct, 2008, 07:05: PM

Expert Answer:

tanAtanB=(a-b/a+b)½

tan2 A tan2 B=(a-b/a+b)

Hence:

a/b=(1+tan2 A tan2 B)/(1-tan2 A tan2 B)

Now:

(a-bcos2A)(a-bcos2B)=b2(a/b-cos2A)(a/b-cos2B)=b2((1+tan2 A tan2 B)/(1-tan2 A tan2 B)-cos2A)/(1+tan2 A tan2 B)/(1-tan2 A tan2 B) -cos2B))  =b2((1+tan2 A tan2 B+1-tan2 A tan2 B-2cos2 A+2sin2 A tan2 B)/(1+tan2 A tan2 B+1-tan2 A tan2 B-2cos2 B+2sin2 B tan2 A)

On simplifying:

=b2 ((1+tan2 A tan2 B)/(1-tan2 A tan2 B))2 -1)=b2 (a2 /b2 -1)=a2 -b2

Answered by  | 15th Jun, 2009, 12:49: AM

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