Question

(cos20+8sin10sin50sin70)/sin^2(80)

Answer 1

Useful identity: sinϴsin(60-ϴ)sin(60+ϴ)=1/4*sin3ϴ

So, (cos20+8sin10sin50sin70)/sin^2(80)

= cos20+8(1/4*sin3(10))/sin^2(80)

=cos20+1/sin^2(80)

=2cos^2(10)/sin^2(80)

=2sin^2(80)/sin^2(80)

=2

Answer 2

Given:

$\frac {(cos 20+8 sin 10 sin 50 sin 70)}{sin^2 {80}}$

To find:

The value of $\frac {(cos 20+ 8 sin 10 sin 50 sin 70)}{sin^2 {80}}$

Answer:

Identity:$sin \theta sin (60- \theta) sin (60+\theta )=\frac {1}{4} \times {sin 3 \times \theta}$

So,$\frac {cos 20+ 8 sin 10 sin 50 sin 70)}{sin^2 {80}}$

$= cos 20+8( \frac {1}{4} \times \frac {sin 3 \times {10}}{sin^2{80}}$

$=cos 20+\frac {1}{sin^2{80}}$

$=2 \times \frac {cos^2 {10}}{sin^2 {80}}$

$=2 \times sin^2 {80} {sin^2 {80}}$

=2