In a Δ A B C , a cos B + b cos C + c cos A = a + b + c 2 then

sin ( A - B ) + sin ( B - C ) + sin ( C - A ) = 3 2

4 sin A - B 2 sin B - C 2 sin C - A 2 = 1

The smallest positive value of 8 2 π p for which sin ( p cos θ ) = cos ( p sin θ ) has a solution in is

Given sin ( p cos θ ) = cos ( p sin θ ) ⇒ cos ( π 2 - p cos θ ) = cos ( p sin θ ) ⇒ π 2 - p cos θ = 2 n π ± ( p sin θ ) n ∈ I ⇒ π 2 - 2 n π = p cos θ ± sin θ ⇒ cos θ ± sin θ = 1 - 4 n π 2 p ∵ - 2 ≤ cos θ ± sin θ ≤ 2 ⇒ ∵ - 2 ≤ 1 - 4 n π 2 p ≤ 2 ⇒ 1 - 4 n π 2 p ≤ 2 ⇒ 1 - 4 n π 2 2 ≤ p ⇒ The smallest possible value of p is π 2 2 ∴ The smallest possible value of 8 2 π p is 4 .

If sec ( θ - ϕ ) , sec θ and sec ( θ + ϕ ) are in A . P . then cos 2 θ sec 2 ϕ 2 + 3 , is equal to

Given sec ( θ - ϕ ) , sec θ , sec ( θ + ϕ ) are in A.P. ⇒ sec ( θ - ϕ ) + sec ( θ + ϕ ) = 2 sec θ ⇒ cos ( θ - ϕ ) + cos ( θ + ϕ ) cos ( θ + ϕ ) cos ( θ - ϕ ) = 2 cos θ ⇒ 2 cos θ cos ϕ cos ( θ + ϕ ) cos ( θ - ϕ ) = 2 cos θ ⇒ 2 cos θ cos ϕ cos 2 θ + cos 2 ϕ = 1 cos θ ⇒ 2 cos 2 θ cos ϕ = cos 2 θ + cos 2 ϕ ⇒ 2 cos 2 θ cos ϕ = 2 cos 2 θ - 1 + 1 - 2 sin 2 ϕ ⇒ cos 2 θ cos ϕ - cos 2 θ = = - 1 - cos ϕ 1 + cos ϕ ⇒ - cos 2 θ 1 - cos ϕ = - 1 - cos ϕ 1 + cos ϕ ⇒ cos 2 θ = 2 cos 2 ϕ 2 ⇒ cos 2 θ · sec 2 ϕ 2 = 2 ∴ cos 2 θ · sec 2 ϕ 2 + 3 = 5 .

If 4 tan θ tan ϕ = 3 , then cos ( θ - ϕ ) cos ( θ + ϕ ) is equal to

Given 4 tan θ tan ϕ = 3 ⇒ sin θ sin ϕ cos θ cos ϕ = 3 4 ⇒ cos θ cos ϕ sin θ sin ϕ = 4 3 Using compounded and dividend, ⇒ cos θ cos ϕ + sin θ sin ϕ cos θ cos ϕ - sin θ sin ϕ = 4 + 3 4 - 3 Using the C-D formula, ∴ cos ( θ - ϕ ) cos ( θ + ϕ ) = 7