Prove that tan θ + cot θ ≡ 2 sin 2 θ .

To prove: tan θ + cot θ ≡ 2 sin 2 θ LHS : tan θ + cot θ ≡ sin θ cos θ + cos θ sin θ ≡ sin 2 θ + cos 2 θ sin θ cos θ ≡ 1 sin θ cos θ ≡ 2 2 sin θ cos θ ≡ 2 sin 2 θ = RHS

Prove that 2 cosec 2 θ tan θ ≡ sec 2 θ

To prove: 2 cosec 2 θ tan θ ≡ sec 2 θ Consider, LHS : 2 cosec 2 θ tan θ = 2 tan θ sin 2 θ ≡ 2 tan θ 2 sin θ cos θ ≡ sin θ cos θ sin θ cos θ ≡ sin θ sin θ cos 2 θ ≡ 1 cos 2 θ = sec 2 θ = LHS

Prove that cos 4 x + 4 cos 2 x ≡ 8 cos 4 x - 3 . Hence solve the equation 2 cos 4 x + 8 cos 2 x = 3 for - π ⩽ x ⩽ π .

To prove: cos 4 x + 4 cos 2 x ≡ 8 cos 4 x - 3 LHS : cos 4 x + 4 cos 2 x = cos ( 2 × 2 x ) + 4 cos 2 x = 2 cos 2 ( 2 x ) - 1 + 4 cos 2 x = 2 2 cos 2 x - 1 2 - 1 + 4 2 cos 2 x - 1 = 2 4 cos 4 x - 4 cos 2 x + 1 - 1 + 8 cos 2 x - 4 = 8 cos 4 x - 8 cos 2 x + 2 - 1 + 8 cos 2 x - 4 = 8 cos 4 x - 3 = RHS Hence, showed. Now, 2 cos 4 x + 8 cos 2 x = 2 ( cos 4 x + 4 cos 2 x ) ⇒ 2 8 cos 4 x - 3 = 3 ⇒ 2 8 cos 4 x - 3 = 3 ⇒ 8 cos 4 x = 9 2 ⇒ cos 2 x = ± 9 16 = ± 3 4 ⇒ cos x = ± 3 2 ⇒ x = ± π 6 or x = ± 5 π 6