Show that 1+4sin20°sin140° / sin70° = 2
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1
Answer:
2sin10
1−4sin10sin70
[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.
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