# CBSE Class 10 Math Question Paper 2018 | Q19A

∴ 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)})
$\frac{\text{4sinθ - cosθ + 1}}{\text{4sinθ + cosθ - 1}}$ = $\frac{13}{11}$

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