if a cos theta- b sin theta =c show that +- root of a square + b square - c square
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I am considering theta as 't' for the purpose of clarity...
First square both the sides of provided equation.
==> {acost - bsint}^2 = c^2
==> a^2(cost)^2 + b^2(sint)^2 -2abcostsint=c^2---eq(1)
Now consider LHS of the equation which should be proved right.Let that be equal to 't'
y= {asint+bcost}
y^2=a^2(sint)^2 + b^2(cost)^2 + 2absintcost---eq(2)
By adding equations eq(1) and eq(2) we get
a^2{(sint)^2 + (cost)^2} + b^2{ (sint)^2 + (cost)^2}=y^2 + c^2
==> a^2 + b^2 = y^2 + c^2
===>y=+/-
Hence proved
First square both the sides of provided equation.
==> {acost - bsint}^2 = c^2
==> a^2(cost)^2 + b^2(sint)^2 -2abcostsint=c^2---eq(1)
Now consider LHS of the equation which should be proved right.Let that be equal to 't'
y= {asint+bcost}
y^2=a^2(sint)^2 + b^2(cost)^2 + 2absintcost---eq(2)
By adding equations eq(1) and eq(2) we get
a^2{(sint)^2 + (cost)^2} + b^2{ (sint)^2 + (cost)^2}=y^2 + c^2
==> a^2 + b^2 = y^2 + c^2
===>y=+/-
Hence proved
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