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Question

Prove that 1sin103cos10=4

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Solution

L.H.S=1cos103sin10sin10cos10
put 1 = r cos α,3 = r sin α
r2=1+3=4,r=2
and tanα=3,α=60
L.H.S=r(cos60cos10sin60sin10)sin10cos10
=2cos(60+10)12sin20=4cos70sin20=4sin20sin20=4

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