If α + β = π 2 , show that the maximum value of cos ⁡ α cos ⁡ β is 1 2 .

Given α + β = π 2 β = π 2 - α cosβ = cos π 2 - α = sinα _ _ _ _ _ _ _ _ _ _ _ _ _ _ 1 let M = cosα . cosβ = cosα . sinα put cosβ = sinα byequation 1 = 1 2 2 sinα . cosα M = 1 2 sin 2 α M = 1 2 sin 2 α Max m value of M = 1 2 sin 2 α Max m = 1 2 × 1 = 1 2

Prove that tan θ tan 60 ° - θ · tan 60 ° + θ = tan 3 θ

Given, L . H . S . = tan θ tan 60 ° - θ tan 60 ° + θ = tan θ . tan 60 ° - tan θ 1 + tan 60 ° tan θ tan 60 ° + tan θ 1 - tan 60 ° tan θ = tan θ 3 - tan θ 1 + 3 tan θ 3 + tan θ 1 - 3 tan θ = tan θ 3 2 - tan 2 θ 1 - 3 tan θ 2 = tan θ 3 - tan 2 θ 1 - 3 tan 2 θ = 3 tan θ - tan 3 θ 1 - 3 tan 2 θ = tan 3 θ = R . H . S .

If cos α = 1 2 , sin β = 1 3 , show that tan α + β 2 cot α - β 2 = 5 + 2 6 or 5 - 2 6

Given that cos α = 1 2 , sin β = 1 3 Use sin 2 x + cos 2 x = 1 , we get sin α = 1 - 1 2 2 = 1 2 Now, tan α + β 2 cot α - β 2 = sin α + β 2 cos α + β 2 × cos α - β 2 sin α - β 2 ⇒ tan α + β 2 cot α - β 2 = 2 sin α + β 2 cos α - β 2 2 cos α + β 2 sin α - β 2 Use sin A + sin B = 2 sin A + B 2 cos A - B 2 and sin A - sin B = 2 cos A + B 2 sin A - B 2 , we get tan α + β 2 cot α - β 2 = sin α + sin β sin α - sin β ⇒ tan α + β 2 cot α - β 2 = 1 2 + 1 3 1 2 - 1 3 ⇒ tan α + β 2 cot α - β 2 = 3 + 2 3 - 2 × 3 + 2 3 + 2 ⇒ tan α + β 2 cot α - β 2 = 3 + 2 + 2 6 3 - 2 ⇒ tan α + β 2 cot α - β 2 = 5 + 2 6 Hence, tan α + β 2 cot α - β 2 = 5 + 2 6 or 5 - 2 6

Prove that s i n 65 ° + c o s 65 ° = 2 c o s 20 ° .

To prove, sin 65 ° + cos 65 ° = 2 cos 20 ° LHS = sin 65 ° + cos 65 ° = cos 90 ° - 65 ° + cos 65 ° = cos 25 ° + cos 65 ° = 2 cos 90 ° 2 cos 40 ° 2 ∵ cos C + cos D = 2 cos C + D 2 cos C - D 2 = 2 cos 45 ° cos 20 ° = 2 × 1 2 cos 20 ° = 2 cos 20 ° = RHS Hence, proved.