If 4𝑡𝑎𝑛θ = 3 then evaluate, 4sinθ−cosθ+14sinθ+cosθ−1

Solution:

It is given that 4𝑡𝑎𝑛θ = 3

$\begin{aligned} i.e \tan θ = \frac{3}{4} \\ \tan θ = \frac{opposite side of θ}{adjacent side of θ} \end{aligned}$

By hypotenuse rule

$\begin{aligned} A C^{2} = B C^{2} + A B^{2} \\ A C^{2} = 4^{2} + 3^{2} \\ A C^{2} = 16 + 9 \\ A C^{2} = 25 \\ A C = 5 \end{aligned}$

$\begin{aligned} Now, \sin θ = \frac{opposite side of θ}{hypotenuse} \\ \sin θ = \frac{A B}{A C} \\ \sin θ = \frac{3}{5} \\ Cos θ = \frac{adjacent side of θ}{hypotenuse} \end{aligned}$

$\begin{aligned} \cos θ = \frac{4}{5} \\ Now, \frac{4 \sin θ - \cos θ + 1}{4 \sin θ + \cos θ - 1} = \frac{4 (\frac{3}{5}) - \frac{4}{5} + 1}{4 (\frac{3}{5}) + \frac{4}{5} - 1} \\ = \frac{\frac{8}{5} + 1}{\frac{16}{5} - 1} \\ = \frac{\frac{13}{5}}{\frac{11}{5}} \\ = \frac{13}{11} \end{aligned}$

If 4𝑡𝑎𝑛θ = 3 then evaluate, $(\frac{4 \sin θ - \cos θ + 1}{4 \sin θ + \cos θ - 1})$

Solution:

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