Question

If tanβ=1 then, (8sinβ+5cosβ)/(sin³β-2cos³β+7cosβ)=?

Answer 1

Step-by-step explanation:

tan a=1

then

a=45°

then

(8sinβ+5cosβ)/ (sin³β-2cos³β+7cosβ)

=(8sin45°+5cos45°)/(sin³45°-2cos³45°+7cos45°)

=[(8*1/√2)+(5*1/√2)]/[(1/√2)³-2*(1/√2)³+(7*1/√2)]

=[(8/√2)+(5/√2)]/[(1/2√2)-(2/2√2)+(7/√2)]

=(13/√2)/[(-1/2√2)+(14/2√2)]

=(13/√2)/(13/2√2)

=2