Question

If tanalpha equals the integral solution of the inequality 4x^2-16x+15<0 and cosbeta equals to the slope of bisector of the first quadrant, then sin(alpha+beta)sin(alpha-beta) is equal to​

Answer 1

Answer:

Dear student

We have,4x2−16x+15<0(2x−3)(2x−5)<032<x<52⇒tanα=2⇒cotα=12and Given that cosβ=slope bisector of the first quadrant=tan45°=1So, β=0Now, sin(α+β)sin(α−β)=sin2α−sin2β=11+cot2α=11+14=154=45