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Question

Mathematics Question on Trigonometric Equations

Ifα,−π2<α<π2\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}is the solution of4cosθ+5sinθ=1 4cos\theta+ 5sin\theta=1then the value of tanαtan\alpha is

A

10−106\frac{10-\sqrt10}{6}

B

10−1012\frac{10-\sqrt10}{12}

C

10−1012\frac{\sqrt10-10}{12}

D

10−106\frac{\sqrt10-10}{6}

Answer

10−1012\frac{\sqrt10-10}{12}

Explanation

Solution

Step 1. Rewrite the Equation in Terms of tan θ

Given 4 + 5 tan⁡θ\tan θ = sec⁡θ\sec θ.

Step 2. Square Both Sides

Squaring both sides to eliminate sec⁡θ\sec θ , we get:

24tan⁡2θ+40tan⁡θ+15=024 \tan^2 θ + 40 \tan θ + 15 = 0

Step 3. Solve for tan θ

Solving this quadratic equation, we find:

tan⁡θ=−10±1012\tan θ = \frac{-10 \pm \sqrt{10}}{12}

Step 4. Choose the Correct Value Based on Range

Since −π2<α<π2-\frac{\pi}{2} < α < \frac{\pi}{2}, we reject tan⁡α=−10+1012\tan α = \frac{-10 + \sqrt{10}}{12} and select:

tan⁡α=10−1012\tan α = \frac{\sqrt{10} - 10}{12}

So, the correct answer is: 10−1012\frac{\sqrt{10} - 10}{12}