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JEE Main 2018
Trigonometric Equations
Trigonometric Equations
Hard

Question

If α,−π2<α<π2\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2} is the solution of 4cos⁡θ+5sin⁡θ=14 \cos \theta+5 \sin \theta=1, then the value of tan⁡α\tan \alpha is

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Solution

4+5tan⁡θ=sec⁡θ4+5 \tan \theta=\sec \theta Squaring : 24tan⁡2θ+40tan⁡θ+15=024 \tan ^2 \theta+40 \tan \theta+15=0 tan⁡θ=−10±1012\tan \theta=\frac{-10 \pm \sqrt{10}}{12} and tan⁡θ=−(10+1012)\tan \theta=-\left(\frac{10+\sqrt{10}}{12}\right) is Rejected. (3) is correct.

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