Question

1-4 sin10°sin 70°
÷
2 sin 10​

Answer 1

Answer:

[2sinAsinB=cos(A−B)−cos(A+B)]

=

2sin10

1−2[cos(10−70)−cos(10+70)]

[cos(−θ)=cosθ]

=

2sin10

1−2cos60+2cos80

=

2sin10

1−2(1/2)+2cos80

=

sin10

cos80

[cos(90−θ)=sinθ,∴cos(90−10)=sin10]

=

sin10

sin10

=1.