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Trigonometric Identities

If 3 cot θ = ...

Question

If 3 cot $θ = 2$ , show that $\frac{4 s i n θ - 3 c o s θ}{(2 s i n θ + 6 c o s θ)} = \frac{1}{3}$ .

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Solution

$rightwards double arrow 3 c o t theta equals 2 rightwards double arrow c o t theta equals 2 over 3 rightwards double arrow tan theta equals fraction numerator 1 over denominator c o t theta end fraction equals fraction numerator 1 over denominator begin display style bevelled 2 over 3 end style end fraction equals 3 over 2 rightwards double arrow s e c theta equals square root of 1 plus tan squared theta end root equals square root of 1 plus left parenthesis 3 over 2 right parenthesis squared end root equals square root of 1 plus 9 over 4 end root equals fraction numerator square root of 13 over denominator 2 end fraction rightwards double arrow cos theta equals fraction numerator 1 over denominator s e c theta end fraction equals fraction numerator 1 over denominator begin display style bevelled fraction numerator square root of 13 over denominator 2 end fraction end style end fraction equals fraction numerator 2 over denominator square root of 13 end fraction rightwards double arrow cos e c theta equals square root of 1 plus c o t squared theta end root equals square root of 1 plus left parenthesis 2 over 3 right parenthesis squared end root equals square root of 1 plus 4 over 9 end root equals fraction numerator square root of 13 over denominator 3 end fraction rightwards double arrow sin theta equals fraction numerator 1 over denominator cos e c theta end fraction equals fraction numerator 1 over denominator begin display style bevelled fraction numerator square root of 13 over denominator 3 end fraction end style end fraction equals fraction numerator 3 over denominator square root of 13 end fraction rightwards double arrow L H S equals fraction numerator 4 sin theta minus 3 cos theta over denominator 2 sin theta plus 6 cos theta end fraction equals fraction numerator 4 cross times begin display style fraction numerator 3 over denominator square root of 13 end fraction end style minus 3 cross times begin display style fraction numerator 2 over denominator square root of 13 end fraction end style over denominator 2 cross times begin display style fraction numerator 3 over denominator square root of 13 end fraction end style plus 6 cross times begin display style fraction numerator 2 over denominator square root of 13 end fraction end style end fraction equals fraction numerator begin display style bevelled fraction numerator 6 over denominator square root of 13 end fraction end style over denominator begin display style bevelled fraction numerator 18 over denominator square root of 13 end fraction end style end fraction equals 6 over 18 equals 1 third equals R H S$
Hence proved.

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