Question

Show that : [3cos π\3•secπ\3-4sin5π\6•tanπ4]×cos2π =1

Answer 1

Answer:

Step-by-step explanation:

[3cos π/3•secπ/3 - 4sin5π/6•tanπ/4]×cos2π =1

L.H.S

[3Cos60° * 1/Cos60° - 4 Sin150° * Tan45°] *Cos360°

= [3 - 4 Sin150° * 1 ] * 1

= 3 - 4 Sin150°

= 3 - 4Sin(90+60)°

= 3 - 4 Cos60°  ( ∵ Sin(90 + ∅) = Cos∅)

= 3 - 4 * 1/2

= 3 - 2

= 1

= R.H.S

Hence proved.