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If 3 tan theta =4 , prove that 5sin alpha - 3 cos alpha / 3 sin alpha + 3 cis alpha = 11/21​

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We have been given that
3\tan(\alpha)=4. We are asked to prove that
(5\sin(\alpha)-3\cos(\alpha))/(3\sin(\alpha)+3\cos(\alpha))=(11)/(21).

First of all, we will solve for tangent of alpha as:


\tan(\alpha)=(4)/(3)

Now, we will divide left side of our given equation by
\cos(\alpha) as:


((5\sin(\alpha))/(\cos(\alpha))-(3\cos(\alpha))/(\cos(\alpha)))/((3\sin(\alpha))/(\cos(\alpha))+(3\cos(\alpha))/(\cos(\alpha)))

We know that
(\sin(x))/(\cos(x))=\tan(x), so we will get:


(5\tan(\alpha)-3)/(3\tan(\alpha)+3)

Upon substituting
\tan(\alpha)=(4)/(3), we will get:


(5\tan(\alpha)-3)/(3\tan(\alpha)+3)=(5\cdot(4)/(3)-3)/(3\cdot(4)/(3)+3)


(5\tan(\alpha)-3)/(3\tan(\alpha)+3)=((20)/(3)-3)/((12)/(3)+3)


(5\tan(\alpha)-3)/(3\tan(\alpha)+3)=((20)/(3)-(9)/(3))/((12)/(3)+(9)/(3))


(5\tan(\alpha)-3)/(3\tan(\alpha)+3)=((20-9)/(3))/((12+9)/(3))


(5\tan(\alpha)-3)/(3\tan(\alpha)+3)=((11)/(3))/((21)/(3))


(5\tan(\alpha)-3)/(3\tan(\alpha)+3)=(11\cdot3)/(21\cdot 3)


(5\tan(\alpha)-3)/(3\tan(\alpha)+3)=(11)/(21)

Hence proved.

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