If 3 tan θ = 3 sin θ , find the value of sin 2 θ - cos 2 θ

Given that, 3 tan θ = 3 sin θ 3 sin θ cos θ = 3 sin θ ⇒ 3 sin θ cos θ = 3 sin θ ⇒ 1 cos θ = 3 3 = 3 ⇒ cos θ = 1 3 ⇒ cos 2 θ = 1 3 Now, sin 2 θ + cos 2 θ = 1 ⇒ sin 2 θ = 1 - cos 2 θ ⇒ 1 - 1 3 = 1 - 1 3 = 2 3 Therefore, sin 2 θ - cos 2 θ = 2 3 - 1 3 = 1 3 Hence, the required answer is 1 3 .

If sec θ + tan θ = p , prove that sin θ = p 2 - 1 p 2 + 1

Given: sec θ + tan θ = p RHS = x 2 - 1 x 2 + 1 = sec θ + tan θ 2 - 1 sec θ + tan θ 2 + 1 = sec 2 θ + tan 2 θ + 2 tan θsec θ - 1 sec 2 θ + tan 2 θ + 2 tan θsec θ + 1 ∵ a ± b 2 = a 2 ± 2 ab ± b 2 = tan 2 θ + tan 2 θ + 2 tan θsec θ sec 2 θ + sec 2 θ + 2 tan θsec θ = 2 tan 2 θ + 2 tan θsec θ 2 sec 2 θ + 2 tan θsec θ = 2 tan θ tan θ + sec θ 2 sec θ tan θ + sec θ = tan θ sec θ = sin θ = LHS

If tan A = n tan B and sin A = m sin B , prove that cos 2 A = m 2 - 1 n 2 - 1

We have, tan A = n tan B ⇒ tan 2 A = n 2 tan 2 B ⇒ sin 2 A cos 2 A = n 2 × sin 2 B cos 2 B . . . ( i ) Also, sin A = m sin B ⇒ sin 2 A = m 2 sin 2 B ⇒ sin 2 B = sin 2 A m 2 . . . ( i i ) Now, cos 2 B = 1 - sin 2 B = 1 - sin 2 A m 2 [ from ( i i ) ] From ( i ) we get, sin 2 A cos 2 A = n 2 × sin 2 B cos 2 B = n 2 × sin 2 A m 2 1 - sin 2 A m 2 = n 2 × sin 2 A m 2 m 2 - sin 2 A m 2 = n 2 × sin 2 A m 2 - sin 2 A ⇒ sin 2 A cos 2 A = n 2 × sin 2 A m 2 - sin 2 A ⇒ 1 cos 2 A = n 2 × 1 m 2 - sin 2 A = n 2 m 2 - 1 + cos 2 A ⇒ cos 2 A = m 2 - 1 + cos 2 A n 2 ⇒ cos 2 A = m 2 - 1 n 2 + cos 2 A n 2 ⇒ cos 2 A - cos 2 A n 2 = m 2 - 1 n 2 ⇒ cos 2 A 1 - 1 n 2 = m 2 - 1 n 2 ⇒ cos 2 A n 2 - 1 n 2 = m 2 - 1 n 2 ⇒ cos 2 A = m 2 - 1 n 2 - 1 Hence, proved.

If cosec θ - sin θ = m and sec θ - cos θ = n , prove that, m 2 n 2 3 + m n 2 2 3 = 1

Given : cosec θ - sin θ = m and sec θ - cos θ = n L . H . S m 2 n = cosec θ - sin θ 2 sec θ - cos θ m 2 n = 1 - sin 2 θ sin θ 2 1 - cos 2 θ cos θ cosec θ = 1 sin θ sec θ = 1 cos θ m 2 n = cos 2 θ 2 × sin 2 θ sin 2 θ cos θ = cos 4 θ sin 2 θ sin 2 θ cos θ = cos 3 θ sin 2 θ + cos 2 θ = 1 ( m 2 n ) 2 3 = ( cos 3 θ ) 2 3 = cos 2 θ → 1 mn 2 = cosec θ - sin θ sec θ - cos θ 2 = 1 - sin 2 θ sin θ 1 - cos 2 θ cos θ 2 mn 2 = cos 2 θ sin θ × sin 2 θ 2 cos 2 θ = sin 3 θ ( mn 2 ) 2 3 = sin 2 θ → ( 2 ) Adding ( 1 ) and ( 2 ) , we get ( m 2 n ) 2 3 + ( mn 2 ) 2 3 = sin 2 θ + cos 2 θ = 1 [From the identity sin 2 θ + cos 2 θ = 1 ] Hence, proved.