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Mathematics

If 12 cosec θ = 13, find the value of 2 sin θ−3 cos θ4 sin θ−9 cos θ\dfrac{2\text{ sin θ} - 3\text{ cos θ}}{4\text{ sin θ} - 9\text{ cos θ}}.

Trigonometric Identities

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Answer

Given 12 cosec θ = 13

⇒ cosec θ = 1312\dfrac{13}{12}

1 + cot2 θ = cosec2 θ

Putting values we get,

1+cot2 θ=(1312)21+cot2 θ=169144cot2 θ=169144−1cot2 θ=169−144144cot2 θ=25144cot θ=25144cot θ=512.1 + \text{cot}^2\spaceθ = \Big(\dfrac{13}{12}\Big)^2 \\[1em] 1 + \text{cot}^2\spaceθ = \dfrac{169}{144} \\[1em] \text{cot}^2\spaceθ = \dfrac{169}{144} - 1 \\[1em] \text{cot}^2\spaceθ = \dfrac{169 - 144}{144} \\[1em] \text{cot}^2\spaceθ = \dfrac{25}{144} \\[1em] \text{cot }θ = \sqrt{\dfrac{25}{144}} \\[1em] \text{cot } θ = \dfrac{5}{12}.

We need to find the value of 2 sin θ−3 cos θ4 sin θ−9 cos θ\dfrac{2\text{ sin θ} - 3\text{ cos θ}}{4\text{ sin θ} - 9\text{ cos θ}}

Dividing numerator and denominator of above expression by sin θ.

⇒2 sin θ−3 cos θ sin θ4 sin θ−9 cos θsin θ=2−3 cot θ4−9 cot θ=2−3×5124−9×512=2−544−154=8−5416−154=3414=3×44=3.\Rightarrow \dfrac{\dfrac{2\text{ sin θ} - 3\text{ cos θ}}{\text{ sin θ}}}{\dfrac{4\text{ sin θ} - 9\text{ cos θ}}{\text{sin θ}}} \\[1em] = \dfrac{2 - 3\text{ cot θ}}{4 - 9\text{ cot θ}} \\[1em] = \dfrac{2 - 3 \times \dfrac{5}{12}}{4 - 9 \times \dfrac{5}{12}} \\[1em] = \dfrac{2 - \dfrac{5}{4}}{4 - \dfrac{15}{4}} \\[1em] = \dfrac{\dfrac{8 - 5}{4}}{\dfrac{16 - 15}{4}} \\[1em] = \dfrac{\dfrac{3}{4}}{\dfrac{1}{4}} \\[1em] = \dfrac{3 \times 4}{4} \\[1em] = 3.

Hence, the value of the expression is 3.

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