Prove that 1 + sec A sec A = sin 2 A 1 - cos A .

We have to prove 1 + sec A sec A = sin 2 A 1 - cos A . R H S = sin 2 A 1 - cos A = 1 - cos 2 A 1 - cos A ∵ sin 2 A = 1 - cos 2 A = 1 - cos A 1 + cos A 1 - cos A = 1 + cos A = 1 + 1 sec A ∵ cos A = 1 sec A = 1 + sec A sec A = L H S 1 + sec A sec A = sin 2 A 1 - cos A

Prove that 1 sec x - tan x - 1 cos x = 1 cos x - 1 sec x + tan x

LHS, 1 sec x - tan x ) - 1 cos x = sec 2 x - tan 2 x sec x - tan x - 1 cos x = sec x + tan x - 1 cos x = tan x = sec x - sec x + tan x = 1 cos x - sec 2 x - tan 2 x sec x + tan x = 1 cos x - 1 sec x + tan x ( Proved ) Hence, L.H.S=R.H.S (Proved)

Prove that sec A + tan A sec A - tan A . cosec A - 1 cosec A + 1 = 1

Given, to prove sec A + tan A sec A - tan A . cosec A - 1 cosec A + 1 = 1 Taking LHS, L H S = sec A + tan A sec A - tan A . cosec A - 1 cosec A + 1 = 1 cos A + sin A cos A 1 cos A - sin A cos A . 1 sin A - 1 1 sin A + 1 = 1 + sin A 1 - sin A . 1 - sin A 1 + sin A = 1 = R H S Hence, L.H.S=R.H.S (Proved)

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

Given, to prove cos e c θ + c o t θ cos e c θ - c o t θ = 1 + 2 c o t 2 θ + 2 cos e c 2 θ cos θ Taking RHS, R H S = 1 + 2 c o t 2 θ + 2 cos e c 2 θ cos θ = 1 + 2 cos 2 θ sin 2 θ + 2 cos 2 θ sin 2 θ = sin 2 θ + 2 cos 2 θ + 2 cos θ sin 2 θ = 1 + cos 2 θ + 2 cos θ sin 2 θ = cos 2 θ + cos θ + cos θ + 1 1 - cos 2 θ = cos θ ( cos θ - 1 ) + 1 ( cos θ + 1 ) ( 1 + cos θ ) ( 1 - cos θ ) = 1 + cos θ 1 - cos θ = 1 sin θ + cos θ sin θ 1 sin θ - cos θ sin θ = cos e c θ + c o t θ cos e c θ - c o t θ = L H S Hence, R.H.S=L.H.S (Proved)

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

Given, to prove sin θ c o t θ + cos e c θ = 2 + sin θ c o t θ - cos e c θ Taking L.H.S L H S = sin θ c o t θ + cos e c θ = sin θ cos θ sin θ + 1 sin θ = sin 2 θ cos θ + 1 = 1 - cos θ Now taking R.H.S R H S = 2 + sin θ c o t θ - cos e c θ = 2 + sin θ cos θ sin θ - 1 sin θ = 2 + sin 2 θ cos θ - 1 = 2 + 1 - cos 2 θ cos θ - 1 = 2 cos θ - 2 + 1 - cos 2 θ cos θ - 1 = 1 - cos θ 2 1 - cos θ = 1 - cos θ Hence, L.H.S=R.H.S (Proved)

If 7 sin 2 A + 3 cos 2 A = 4 , then prove that tan A = 1 3

Given, 7 sin 2 A + 3 cos 2 A = 4 , then prove that tan A = 1 3 Let us simplify the given equation. 7 sin 2 A + 3 cos 2 A = 4 ⇒ 7 tan 2 A 1 + tan 2 A + 3 1 + tan 2 A = 4 ⇒ 7 tan 2 A + 3 1 + tan 2 A = 4 ⇒ 7 tan 2 A + 3 = 4 + 4 tan 2 A ⇒ tan A = 1 3 Hence, L.H.S=R.H.S (Proved)

If cos θ + sin θ = 2 cos θ , then prove that cos θ - sin θ = 2 sin θ

Given, cos θ + sin θ = 2 cos θ , to prove that cos θ - sin θ = 2 sin θ Using the given equation, we have cos θ + sin θ = √ 2 cos θ ⇒ sin θ = √ 2 cos θ - cos θ ⇒ sin θ = ( √ 2 - 1 ) cos θ ⇒ [ sin θ / ( √ 2 - 1 ) ] = cos θ ⇒ [ sin θ ( √ 2 + 1 ) / ( 2 - 1 ) ] = cos θ ⇒ [ √ 2 sin θ + sin θ ] = cos θ ⇒ cos θ - sin θ = √ 2 sin θ So, we get the required results. Hence, Proved