Prove that 1 + cos A 1 - cos A = cos e c A + c o t A .

1 + cos A 1 - cos A = 1 + cos A 1 - cos A × 1 + cos A 1 + cos A = 1 + cos A 2 1 - cos 2 A = 1 + cos A 2 sin 2 A = 1 + cos A sin A = 1 sin A + cos A sin A = cos e c A + c o t A

Prove that 1 + cos A 1 - cos A = cos e c A + c o t A .

1 + cos A 1 - cos A = 1 + cos A 1 - cos A × 1 + cos A 1 + cos A = 1 + cos A 2 1 - cos 2 A = 1 + cos A 2 sin 2 A = 1 + cos A sin A = 1 sin A + cos A sin A = cos e c A + c o t A

If cos e c θ + c o t θ = p , prove that cos θ = p 2 - 1 p 2 + 1

R.H.S. = p 2 - 1 p 2 + 1 = cos e c θ + c o t θ 2 - 1 cos e c θ + c o t θ 2 + 1 = cos e c 2 θ + c o t 2 θ + 2 cos e c θ c o t θ - 1 cos e c 2 θ + c o t 2 θ + 2 cos e c θ c o t θ + 1 = cos e c 2 θ - 1 + c o t 2 θ + 2 cos e c θ c o t θ cos e c 2 θ + 2 cos e c θ c o t θ + c o t 2 θ + 1 = c o t 2 θ + c o t 2 θ + 2 cos e c θ c o t cos e c 2 θ + 2 cos e c θ c o t θ + cos e c 2 θ = 2 c o t 2 θ + 2 cos e c θ c o t θ 2 cos e c 2 θ + 2 cos e c θ c o t θ = 2 c o t θ c o t θ + cos e c θ 2 cos e c θ cos e c θ + c o t θ = c o t θ cos e c θ = cos θ sin θ 1 sin θ = cos θ = L.H.S.

Prove that cos e c A - c o t A 2 = 1 - cos A 1 + cos A .

cos e c A - c o t A 2 = 1 sin A - cos A sin A 2 = 1 - cos A sin A 2 = 1 - cos A 2 sin 2 A = 1 - cos A 2 1 - cos 2 A = 1 - cos A 2 1 - cos A 1 + cos A = 1 - cos A 1 + cos A Thus proved.

If 5 c o t θ = 3 , show that the value of 5 sin θ - 3 cos θ 4 sin θ + 3 cos θ is 16 29 .

Given, 5 c o t θ = 3 ⇒ c o t θ = 3 5 Now, 5 sin θ - 3 cos θ 4 sin + 3 cos θ = 5 sin θ sin θ - 3 cos θ sin θ 4 sin θ sin θ + 3 cos θ sin θ = 5 - 3 c o t θ 4 + 3 c o t θ = 5 - 3 × 3 5 4 + 3 × 3 5 = 5 - 9 5 4 + 9 5 = 16 5 29 5 = 16 29

Prove that 1 + sin θ 1 - sin θ = sec θ + tan θ 2 .

1 + sin θ 1 - sin θ = 1 + sin θ 1 - sin θ × 1 + sin θ 1 + sin θ = 1 + sin θ 2 1 - sin 2 θ = 1 + sin θ 2 cos 2 θ = 1 + sin θ cos θ 2 = 1 cos θ + sin θ cos θ 2 = sec θ + tan θ 2

Prove that 1 sec θ - tan θ - 1 cos θ = 1 cos θ - 1 sec θ + tan θ

L.H.S. = 1 sec θ - tan θ - 1 cos θ = 1 sec θ - tan θ × sec θ + tan θ sec θ + tan θ - 1 cos θ = sec θ + tan θ sec 2 θ - tan 2 θ - sec θ = sec θ + tan θ - sec θ = tan θ . . . i R.H.S. = 1 cos θ - 1 sec θ + tan θ = 1 cos θ - 1 sec θ + tan θ × sec θ - tan θ sec θ - tan θ = 1 cos θ - sec θ - tan θ sec 2 θ - tan 2 θ = sec θ - sec θ + tan θ = tan θ . . . . i i From (i) and (ii), L.H.S. = R.H.S.

Prove that sin A - 2 sin 3 A 2 cos A - cos A = tan A .

L.H.S. = sin A - 2 sin 3 A 2 cos 3 A - cos A = sin A 1 - 2 sin 2 A cos A 2 cos 2 A - 1 = sin A 1 - 2 sin 2 A cos A 2 1 - sin 2 A - 1 = sin A 1 - 2 sin 2 A cos A 2 - 2 sin 2 A - 1 = sin A 1 - 2 sin 2 A cos A 1 - 2 sin 2 A = sin A cos A = tan A = R.H.S.