Prove that 1 + cos ⁡ A sin ⁡ A = sin ⁡ A 1 - cos ⁡ A .

By taking L H S we get, L H S = 1 + cos ⁡ A sin ⁡ A Now, multiplying the numerator and the denominator of L H S by 1 - cos A , we get = ( 1 + cos ⁡ A ) sin ⁡ A × ( 1 - cos ⁡ A ) ( 1 - cos ⁡ A ) Further, using the identity, a 2 - b 2 = a + b a - b , we get, = 1 - cos 2 ⁡ A sin ⁡ A 1 - cos ⁡ A = sin 2 ⁡ A sin ⁡ A 1 - cos ⁡ A ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 = sin ⁡ A 1 - cos ⁡ A = R H S Hence, proved.

Prove that 1 + cos ⁡ A sin ⁡ A = sin ⁡ A 1 - cos ⁡ A .

By taking L H S we get, L H S = 1 + cos ⁡ A sin ⁡ A Now, multiplying the numerator and the denominator of L H S by 1 - cos A , we get = ( 1 + cos ⁡ A ) sin ⁡ A × ( 1 - cos ⁡ A ) ( 1 - cos ⁡ A ) Further, using the identity, a 2 - b 2 = a + b a - b , we get, = 1 - cos 2 ⁡ A sin ⁡ A 1 - cos ⁡ A = sin 2 ⁡ A sin ⁡ A 1 - cos ⁡ A ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 = sin ⁡ A 1 - cos ⁡ A = R H S Hence, proved.

Prove that, 1 + sin ⁡ A 1 - sin ⁡ A = sec ⁡ A + tan ⁡ A .

By taking LHS, L H S = 1 + sin ⁡ A 1 - sin ⁡ A On rationalizing the LHS, we get 1 + sin ⁡ A 1 - sin ⁡ A × 1 + sin ⁡ A 1 + sin ⁡ A = 1 + sin ⁡ A 2 1 - sin 2 ⁡ A ∵ a + b a - b = a 2 - b 2 = 1 + sin ⁡ A 2 cos 2 ⁡ A ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 = 1 cos ⁡ A + sin ⁡ A cos ⁡ A = sec ⁡ A + tan ⁡ A ∵ 1 cos ⁡ θ = s e c θ , sin ⁡ θ cos ⁡ θ = tan ⁡ θ = R H S Hence, proved.

Prove that: 1 - cos ⁡ A 1 + cos ⁡ A + 1 + cos ⁡ A 1 - cos ⁡ A = 2 cosec ⁡ A .

Given equation is 1 - cos ⁡ A 1 + cos ⁡ A + 1 + cos ⁡ A 1 - cos ⁡ A = 2 cosec ⁡ A Simplifying the LHS 1 - cos ⁡ A 1 + cos ⁡ A + 1 + cos ⁡ A 1 - cos ⁡ A Taking LCM of denominators, = 1 - cos A 1 - cos A + 1 + cos A 2 1 + cos A 1 - cos A = ( 1 - cos A ) 2 + ( 1 + cos A ) 1 - cos 2 A = 1 - cos A + 1 + cos A 1 - cos 2 A ∵ sin 2 A + cos 2 A = 1 = 2 sinA = 2 cosecA = RHS Hence, proved.

Prove that, sec θ - 1 sec θ + 1 + sec θ + 1 sec θ - 1 = 2 cosec θ .

By taking LHS, we get LHS = sec θ - 1 sec θ + 1 + sec θ + 1 sec θ - 1 = sec θ - 1 sec θ + 1 + sec θ + 1 sec θ - 1 = sec θ - 1 sec θ - 1 + sec θ + 1 sec θ + 1 sec θ + 1 sec θ - 1 = ( sec θ - 1 ) 2 + ( sec θ + 1 ) 2 sec 2 θ - 1 = sec θ - 1 + sec θ + 1 sec 2 θ - 1 = 2 sec θ tan 2 θ = 2 sec θ tan θ ∵ sec 2 θ - tan 2 θ = 1 = 2 1 cos θ sin θ cos θ ∵ sec θ = 1 cos θ , tan θ = sin θ cos θ = 2 sin θ = 2 c o s e c θ [ ∵ cosec θ = 1 sin θ ] = R H S Hence, proved.

Prove that, 1 + sin ⁡ θ 1 - sin ⁡ θ = 1 - sin ⁡ θ 1 + sin ⁡ θ = 2 sec ⁡ θ .

Given equation is: 1 + sin ⁡ θ 1 - sin ⁡ θ = 1 - sin ⁡ θ 1 + sin ⁡ θ = 2 sec ⁡ θ By taking LHS, we get, 1 + sin ⁡ θ 1 - sin ⁡ θ + 1 - sin ⁡ θ 1 + sin ⁡ θ = 1 + sin ⁡ θ 1 - sin ⁡ θ + 1 - sin ⁡ θ 1 + sin ⁡ θ = 1 + sin ⁡ θ 2 + 1 - sin ⁡ θ 2 1 - sin ⁡ θ 1 + sin ⁡ θ = 1 + sin ⁡ θ + 1 - sin ⁡ θ 1 - sin 2 ⁡ θ ∵ a + b a - b = a 2 - b 2 = 2 cos 2 ⁡ θ ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 = 2 cos ⁡ θ = 2 sec ⁡ θ [ ∵ sec ⁡ θ = 1 cos ⁡ θ ] = R H S Hence, proved.

Prove that, 1 + cos ⁡ θ 1 - cos ⁡ θ + 1 - cos ⁡ θ 1 + cos ⁡ θ = 2 cosec ⁡ θ .

Given equation is: 1 + cos ⁡ θ 1 - cos ⁡ θ + 1 - cos ⁡ θ 1 + cos ⁡ θ = 2 cosec ⁡ θ Taking the LHS of the equation, 1 + cos ⁡ θ 1 - cos ⁡ θ + 1 - cos ⁡ θ 1 + cos ⁡ θ = 1 + cos ⁡ θ 1 - cos ⁡ θ + 1 - cos ⁡ θ 1 + cos ⁡ θ = 1 + cos ⁡ θ 1 + cos ⁡ θ + 1 - cos ⁡ θ 1 - cos ⁡ θ 1 - cos ⁡ θ 1 + cos ⁡ θ = 1 + cos ⁡ θ 2 + 1 - cos ⁡ θ 2 1 - cos ⁡ θ 1 + cos ⁡ θ = 1 + cos ⁡ θ + 1 - cos ⁡ θ 1 - cos 2 ⁡ θ ∵ a + b a - b = a 2 - b 2 = 2 sin 2 ⁡ θ ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 = 2 sin ⁡ θ = 2 c o s e c θ = R H S Hence, proved.

Prove that, sec ⁡ θ - 1 sec ⁡ θ + 1 = sec ⁡ θ 1 + cos ⁡ θ 2 .

LHS = sec ⁡ θ - 1 sec ⁡ θ + 1 = 1 cos ⁡ θ - 1 1 cos ⁡ θ + 1 [ ∵ sec ⁡ θ = 1 cos ⁡ θ ] = 1 - cos ⁡ θ cos ⁡ θ 1 + cos ⁡ θ cos ⁡ θ = 1 - cos ⁡ θ 1 + cos ⁡ θ Multiplying both the numerator and the denominator by 1 + cos ⁡ θ , we get = 1 - cos ⁡ θ 1 + cos ⁡ θ 1 + cos ⁡ θ 1 + cos ⁡ θ = 1 - cos 2 ⁡ θ 1 + cos ⁡ θ 2 ∵ a + b a - b = a 2 - b 2 = sin 2 ⁡ θ 1 + cos ⁡ θ 2 ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 = sin ⁡ θ 1 + cos ⁡ θ 2 = R H S Hence, proved.

Prove that, sec ⁡ A - tan ⁡ A 2 = 1 - sin ⁡ A 1 + sin ⁡ A .

LHS = sec ⁡ A - tan ⁡ A 2 Now, using sec ⁡ θ = 1 cos ⁡ θ and tan ⁡ θ = sin ⁡ θ cos ⁡ θ , we get = 1 cos ⁡ A - sin ⁡ A cos ⁡ A 2 = 1 - sin ⁡ A cos ⁡ A 2 = 1 - sin ⁡ A 2 cos ⁡ A 2 = 1 - sin ⁡ A 2 1 - sin 2 ⁡ A ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 = 1 - sin ⁡ A 2 1 - sin ⁡ A 1 + sin ⁡ A ∵ a + b a - b = a 2 - b 2 = 1 - sin ⁡ A 1 + sin ⁡ A = R H S Hence, proved.

Prove that , 1 - cos ⁡ A 1 + cos ⁡ A = cot ⁡ A - cosec ⁡ A 2

LHS = 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 ∵ a + b a - b = a 2 - b 2 = 1 + cos 2 ⁡ A - 2 cos ⁡ A sin 2 ⁡ A ∵ a - b 2 = a 2 - 2 a b + b 2 , sin 2 θ + cos 2 θ = 1 = 1 sin 2 ⁡ A + cos 2 ⁡ A sin 2 ⁡ A - 2 cos ⁡ A sin 2 ⁡ A Using c o s e c θ = 1 sin ⁡ θ and cot ⁡ θ = cos ⁡ θ sin ⁡ θ , we get = c o s e c 2 A + cot 2 ⁡ A - 2 cot ⁡ A c o s e c A = cot ⁡ A - c o s e c A 2 [ ∵ ( a + b ) 2 = a 2 + 2 a b + b 2 ] = RHS Hence, proved.