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if the value of asintheta +bcostheta=c then the value of acostheta-bcostheta is
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first square both the sides then convert sintheta to costheta and solve
\[(a \sin \theta)^2+(b \cos \theta)^2=c^2\]
(asinx+bcosx)^2=c^2
no squaring is not done like that you have to do the whole square
\[(asin \theta+bcos \theta)^{2}\]
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a^2sin^2x+b^2cos^2x+2asinxbcox=c^2
for convenience i am using x
yup like that, and then change sintheta into costheta and solve
a-acos^2x+bcos^2x+2absinxcosx=c^2 a-cos^2x(a+b)+2absinxcosx=c^2
sorry in my question acostheta-bsinthetha to find
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i am telling u wait.
i have solved in the attachment
i have many doubts
you have seen the attachment
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