if tanA+sinA=m. and tanA-sinA=n then find sinA*cosA
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tanA+sinA=m _____(1)
tanA-sinA=n _____(2)
Adding (1) & (2), tanA = (m+n)/2
Subtracting (1) & (2), sinA = (m-n)/2
cosA = sinA / tanA = (m-n)/(m+n)
Thus, sinA × cosA = (m-n)^2 / [2×(m+n)]
tanA-sinA=n _____(2)
Adding (1) & (2), tanA = (m+n)/2
Subtracting (1) & (2), sinA = (m-n)/2
cosA = sinA / tanA = (m-n)/(m+n)
Thus, sinA × cosA = (m-n)^2 / [2×(m+n)]
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