Prove that 2 sin 2 π 6 + cosec 7 π 6 cos 2 π 3 = 0 .

LHS = 2 sin 2 π 6 + cosec 7 π 6 cos 2 π 3 = 2 1 2 2 + cosec π + π 6 1 2 2 = 1 2 - 1 4 cosec π 6 ∵ cosec ( π + θ ) = - cosec θ = 1 2 - 1 4 · 2 = 1 2 - 1 2 = 0 = RHS Hence, proved that 2 sin 2 π 6 + cosec 7 π 6 cos 2 π 3 = 0 .

Prove that 3 cos 2 π 4 + sec 2 π 3 + 5 tan 2 π 3 = 29 2 .

LHS = 3 cos 2 π 4 + sec 2 π 3 + 5 tan 2 π 3 = 3 1 2 2 + sec π - π 3 + 5 ( 3 ) 2 = 3 2 - sec π 3 + 5 ( 3 ) ∵ sec ( π - θ ) = - secθ = 3 2 - 2 + 15 = 29 2 = RHS Hence, proved that 3 cos 2 π 4 + sec 2 π 3 + 5 tan 2 π 3 = 29 2 .

Prove that 2 cos 2 π 4 + 2 sin 2 3 π 4 + 2 sec 2 π 3 = 10 .

LHS = 2 cos 2 π 4 + 2 sin 2 3 π 4 + 2 sec 2 π 3 = 2 1 2 2 + 2 sin π - π 4 2 + 2 ( 2 ) 2 = 2 2 + 2 sin 2 π 4 + 8 ∵ sin ( π - θ ) = sinθ = 9 + 2 1 2 = 10 = RHS Hence, proved that 2 cos 2 π 4 + 2 sin 2 3 π 4 + 2 sec 2 π 3 = 10 .

Prove that cot 2 π 6 + cosec 5 π 6 + 3 tan 2 π 6 = 6 .

LHS = cot 2 π 6 + cosec 5 π 6 + 3 tan 2 π 6 = ( 3 ) 2 + cosec π - π 6 + 3 1 3 2 = 3 + cosec π 6 + 3 3 ∵ cosec ( π - θ ) = cosec θ = 4 + 2 = 6 = RHS Hence, proved that cot 2 π 6 + cosec 5 π 6 + 3 tan 2 π 6 = 6 .

Prove that 3 cos π 3 sec π 3 - 4 sin 5 π 6 tan π 4 cos 2 π = 1 .

LHS = 3 cos π 3 sec π 3 - 4 sin 5 π 6 tan π 4 cos 2 π = 3 · 1 2 · 2 - 4 sin π - π 6 · 1 · 1 ∵ cos 2 n π = 1 , n ∈ Z = 3 - 4 sin π 6 ∵ sin ( π - θ ) = sin θ = 3 - 4 · 1 2 = 1 = RHS Hence, proved that 3 cos π 3 sec π 3 - 4 sin 5 π 6 tan π 4 cos 2 π = 1 .