hint:
(m-n)tan(A+B)
= (m-n) (tanA+tanB)/(1-tanAtanB)
= (mtanA - ntanB)/(1-tanAtanB)
now multiply top and bottom by cosAcosB:
= (msinAcosB-nsinBcosA)/(cosAcosB-sinAsinB)
the rest should follow without difficulty.
ifm sin B=nsin(2A+B) prove that(m+n)tanA=(m-n)tan(A+B)
4 answers
Sum
What's this answer
It's really unhelpful
Do they really know the answer or they kept something for simply expecting ( like) from us
It's really unhelpful
Do they really know the answer or they kept something for simply expecting ( like) from us
Use the componendo- dividendo rule then you will get the answer
4 answers