If tan theta equals to a upon B prove that a sin theta + b cos theta upon sin theta minus B cos theta equals to a square minus b square upon a square + b square
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i am taking theta=x
tanx = a/b
(asinx+bcosx)/(asinx-bcosx)
divide the numerator and denominator with cosx.
{(asinx+bcosx)/cosx}/{(asinx-bcosx)/cosx}
(atanx +b)/(atanx-b)
Now put the value of tanx here.
(a(a/b)+b)/(a(a/b)-b)
= (a²+b²)/(a²-b²) ≠ (a²-b²)/(a²+b²) as asked in the question.
I hope you understand the solution.
tanx = a/b
(asinx+bcosx)/(asinx-bcosx)
divide the numerator and denominator with cosx.
{(asinx+bcosx)/cosx}/{(asinx-bcosx)/cosx}
(atanx +b)/(atanx-b)
Now put the value of tanx here.
(a(a/b)+b)/(a(a/b)-b)
= (a²+b²)/(a²-b²) ≠ (a²-b²)/(a²+b²) as asked in the question.
I hope you understand the solution.
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54
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