If-tan-equals-the-integral-solution-of-the-inequality-4x-2-16x-15-lt-0-and-cos-equals-to-the-slope-of-the-bisector-of-the-first-quadrant-then-sin-sin-is-equal-to-

Question Number 12169 by indreshpatelindresh@435gmail.i last updated on 15/Apr/17

If \tan α equals the integral solution of

the inequality 4 x^{2} - 16 x + 15 < 0 and

\cos β equals to the slope of the bisector

of the first quadrant, then

\sin (α + β) \sin (α - β) is equal to

Answered by mrW1 last updated on 15/Apr/17

4 x^{2} - 16 x + 15 = 4 (x^{2} - 4 x + 4) - 1

= 4 {(x - 2)}^{2} - 1 = (2 x - 3) (2 x - 5) = 0

x_{1} = \frac{3}{2} = 1.5

x_{2} = \frac{5}{2} = 2.5

\Rightarrow \tan α = 2

\cos β = \tan 45 ° = 1

\Rightarrow \sin β = 0

\sin (α + β) \sin (α - β)

= (\sin α \cos β + \cos α \sin β) (\sin α \cos β - \cos α \sin β)

= (\sin α) (\sin α)

= \sin^{2} α = \frac{\tan^{2} α}{1 + \tan^{2} α} = \frac{2^{2}}{1 + 2^{2}} = \frac{4}{5}

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