CBSE Class 10 Math Question Paper 2018 | Q19A

Question 19A: If 4 tanθ = 3, evaluate \\frac{\text{4sinθ - cosθ + 1}}{\text{4sinθ + cosθ - 1}}).

Video Explanation

Explanatory Answer | CBSE 2018 Board Paper Q19A

Step 1: Compute the three sides of the right triangle

4tan θ = 3
∴ tan θ = \\frac{3}{4})
tan θ = \\frac{3}{4}) = \\frac{\text{opp side}}{\text{adj side}}) (opposite side 3K, adjacent side 4K)
∴ Hypotenuse = \\sqrt{3k^2 + 4k^2}\\) = 5k

Step 2: Compute values of sin θ and cos θ

sin θ = \\frac{\text{opposite side}}{\text{hypotenuse}}) = \\frac{3k}{5k}) = \\frac{3}{5})
cos θ = \\frac{\text{adjacent side}}{\text{hypotenuse}}) = \\frac{4k}{5k}) = \\frac{4}{5})

Step 3: Compute the value of the expression

\\frac{\text{4sinθ - cosθ + 1}}{\text{4sinθ + cosθ - 1}}) = \\frac{4 × (3/5) - (4/5)) + 1}{4 × (3/5) + (4/5)) - 1}) = \\frac{(12/5)) - (4/5)) + (5/5)}{(12/5)) + (4/5)) - (5/5)}) = \\frac{(13/5)}{(11/5)})
\\frac{\text{4sinθ - cosθ + 1}}{\text{4sinθ + cosθ - 1}}) = \\frac{13}{11})