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Question: If \(\tan\alpha\) equals the integral solution of the inequality \(4x^{2} - 16x + 15 < 0\) and \(\co...

If tan⁡α\tan\alpha equals the integral solution of the inequality 4x2−16x+15<04x^{2} - 16x + 15 < 0 and cos⁡β\cos\beta equals to the slope of the bisector of first quadrant, then sin⁡(α+β)sin⁡(α−β)\sin(\alpha + \beta)\sin(\alpha - \beta) is equal to

A

35\frac{3}{5}

B

−35- \frac{3}{5}

C

25\frac{2}{\sqrt{5}}

D

45\frac{4}{5}

Answer

45\frac{4}{5}

Explanation

Solution

We have 4x2−16x+15<0⇒32<x<524x^{2} - 16x + 15 < 0 \Rightarrow \frac{3}{2} < x < \frac{5}{2}

∴\therefore Integral solution of 4x2−16x+15<04x^{2} - 16x + 15 < 0 is x = 2.

Thus tan⁡α=2.\tan\alpha = 2. It is given that cos⁡β=tan⁡45o=1\cos\beta = \tan 45^{o} = 1

∴sin⁡(α+β)sin⁡(α−β)=sin⁡2α−sin⁡2β\therefore\sin(\alpha + \beta)\sin(\alpha - \beta) = \sin^{2}\alpha - \sin^{2}\beta

=11+cot⁡2α−(1−cos⁡2β)=11+14−0=45= \frac{1}{1 + \cot^{2}\alpha} - (1 - \cos^{2}\beta) = \frac{1}{1 + \frac{1}{4}} - 0 = \frac{4}{5}.