tan^-1(a/b+c)+tan^-1(b/c+a)=π/4. how to prove it?
Answers
Answered by
3
Let 12arccos(ab)=y
Therefore, the required equation becomes
1+tan(y)1−tan(y)+1−tan(y)1+tan(y)
Simplifying the above expression, we get,
(1+tan(y))2+(1−tan(y))21−tan2(y)
=2sec2(y)1−tan2(y)
Dividing both numerator and denominator by sec2(y) we get,
2cos2(y)−sin2(y)
=2cos(2y)
Now,
cos(2y)=cos(arccos(ab))
cos(2y)=ab
Inputting the value of cos(2y) in our obtained expression, we get the required answer.
Hence, the required equation holds true.
Therefore, the required equation becomes
1+tan(y)1−tan(y)+1−tan(y)1+tan(y)
Simplifying the above expression, we get,
(1+tan(y))2+(1−tan(y))21−tan2(y)
=2sec2(y)1−tan2(y)
Dividing both numerator and denominator by sec2(y) we get,
2cos2(y)−sin2(y)
=2cos(2y)
Now,
cos(2y)=cos(arccos(ab))
cos(2y)=ab
Inputting the value of cos(2y) in our obtained expression, we get the required answer.
Hence, the required equation holds true.
Xayro:
thanks
Answered by
5
Let 12arccos(ab)=y
Therefore, the required equation becomes
1+tan(y)1−tan(y)+1−tan(y)1+tan(y)
Simplifying the above expression, we get,
(1+tan(y))2+(1−tan(y))21−tan2(y)
=2sec2(y)1−tan2(y)
Dividing both numerator and denominator by sec2(y) we get,
2cos2(y)−sin2(y)
=2cos(2y)
Now,
cos(2y)=cos(arccos(ab))
cos(2y)=ab
Inputting the value of cos(2y) in our obtained expression, we get the required answer.
Hence, the required equation holds true.
Therefore, the required equation becomes
1+tan(y)1−tan(y)+1−tan(y)1+tan(y)
Simplifying the above expression, we get,
(1+tan(y))2+(1−tan(y))21−tan2(y)
=2sec2(y)1−tan2(y)
Dividing both numerator and denominator by sec2(y) we get,
2cos2(y)−sin2(y)
=2cos(2y)
Now,
cos(2y)=cos(arccos(ab))
cos(2y)=ab
Inputting the value of cos(2y) in our obtained expression, we get the required answer.
Hence, the required equation holds true.
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