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

Given, cos e c θ - c o t θ 2 = 1 - cos θ 1 + cos θ L H S = ( cosec θ - cot θ ) 2 = 1 sin θ - cos θ sin θ 2 [ ∵ cosec θ = 1 sin θ , cot θ = cos θ sin θ ] = 1 - cos θ sin θ 2 = 1 - cos θ 2 sin 2 θ = ( 1 - cos θ ) 2 1 - cos 2 θ [ ∵ sin 2 θ = 1 - cos 2 θ ] = ( 1 - cos θ ) 2 ( 1 - cos θ ) ( 1 + cos θ ) = 1 - cos θ 1 + cos θ = R H S . Hence, proved.

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

Given, cos e c θ - c o t θ 2 = 1 - cos θ 1 + cos θ L H S = ( cosec θ - cot θ ) 2 = 1 sin θ - cos θ sin θ 2 [ ∵ cosec θ = 1 sin θ , cot θ = cos θ sin θ ] = 1 - cos θ sin θ 2 = 1 - cos θ 2 sin 2 θ = ( 1 - cos θ ) 2 1 - cos 2 θ [ ∵ sin 2 θ = 1 - cos 2 θ ] = ( 1 - cos θ ) 2 ( 1 - cos θ ) ( 1 + cos θ ) = 1 - cos θ 1 + cos θ = R H S . Hence, proved.

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

Given, 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.

If c o t A = 8 15 ,then find the value of sin A .

Given that, c o t A = 8 15 We know that, in a right angled triangle, c o t A = B a s e H e i g h t Using Pythagoras theorem, H y p o t e n u s e 2 = B a s e 2 + H e i g h t 2 H y p o t e n u s e 2 = 8 2 + 15 2 = 64 + 225 = 289 H y p o t e n u s e = 289 = 17 sin A = H e i g h t H y p o t e n u s e = 15 17 Hence, sin A = 15 17

Prove that, cos A 1 + sin A + 1 + sin A cos A = 2 sec A

Given: cos A 1 + sin A + 1 + sin A cos A = 2 sec A L H S = cos 2 A + 1 + sin A 2 cos A 1 + sin A = cos 2 A + sin 2 A + 1 + 2 sin A cos A 1 + sin A = 2 + 2 sin A cos A 1 + sin A ∵ sin 2 A + cos 2 A = 1 = 2 1 + sin A cos A 1 + sin A = 2 cos A = 2 sec A ∵ 1 cos A = sec A = R H S Hence, proved.