Prove that: cosec θ - cot θ 2 = 1 - cos θ 1 + cos θ

Let us consider L.H.S: cos e c θ - c o t θ 2 We know that, c o t θ = cos θ sin θ ; cos e c θ = 1 sin θ ⇒ 1 sin θ - cos θ sin θ 2 ⇒ 1 - cos θ sin θ 2 ⇒ 1 - cos θ 2 sin 2 θ We know that, cos 2 θ + sin 2 θ = 1 ⇒ 1 - cos θ 2 1 - cos 2 θ By using the law: a 2 - b 2 = a + b a - b ⇒ 1 - cos θ 1 - cos θ 1 + cos θ 1 - cos θ ⇒ 1 - cos θ 1 + cos θ = R . H . S ∴ cos e c θ - c o t θ 2 = 1 - cos θ 1 + cos θ Hence, proved.

Show that: tan 48 ° tan 23 ° tan 42 ° tan 67 ° = 1

We need to prove that tan 48 ° tan 23 ° tan 42 ° tan 67 ° = 1 We know that tan 90 - θ = cot θ Here tan 42 ° = tan 90 - 48 ° = cot 48 ° tan 67 ° = tan 90 - 23 ° = cot 23 ° We know that, tan θ . cot θ = 1 tan 48 ° tan 23 ° tan 42 ° tan 67 ° can be expressed as tan 48 ° tan 23 ° cot 48 ° cot 23 ° Hence, tan 48 ° tan 23 ° tan 42 ° tan 67 ° = 1

Prove that: cosec θ - cot θ 2 = 1 - cos θ 1 + cos θ

Let us consider L.H.S: cos e c θ - c o t θ 2 We know that, c o t θ = cos θ sin θ ; cos e c θ = 1 sin θ ⇒ 1 sin θ - cos θ sin θ 2 ⇒ 1 - cos θ sin θ 2 ⇒ 1 - cos θ 2 sin 2 θ We know that, cos 2 θ + sin 2 θ = 1 ⇒ 1 - cos θ 2 1 - cos 2 θ By using the law: a 2 - b 2 = a + b a - b ⇒ 1 - cos θ 1 - cos θ 1 + cos θ 1 - cos θ ⇒ 1 - cos θ 1 + cos θ = R . H . S ∴ cos e c θ - c o t θ 2 = 1 - cos θ 1 + cos θ Hence, proved.

Prove that: cosec θ - cot θ 2 = 1 - cos θ 1 + cos θ

We have 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 . Proved.