(1 - cos2 A) cosec2 A = 1
(1 - cos2 A)cosec2 A
= Sin2 A cosec2 A
= (Sin A cosec A)2
= (Sin A × (1/Sin A))2
= (1)2
= 1
(1 + Cot2 A) Sin2 A = 1
We know,
cosec2A - Cot 2 A = 1
So, (1 + Cot2 A) Sin2 A
= Cosec2 A Sin2 A
= (Cosec A Sin A)2
= ((1/Sin A) × Sin A)2
= (1)2
= 1
tan2θ cos2θ = 1 − cos2θ
We know,
sin2 θ + cos2 θ = 1
So, tan2 θ cos2 θ
= (tan θ × cos θ)2
= (sin θ)2 sin2θ 1 - cos2θ
We know,
sin2 θ + cos2 θ = 1
= 1
(sec2θ − 1)(cosec2θ − 1) = 1
We know that, (sec2θ − tan2θ) = 1 (cosec2θ − cot2θ) = 1
So, (sec2θ - 1)(cosec2θ - 1)
= tan2θ × cot2θ
= (tan θ × cot θ)2
= 12
= 1
We know that, (sec2θ − tan2θ) = 1
So,
We know, sin2θ + cos2θ = 1
So, Multiplying both numerator and denominator by (1+ sin θ), we have
We know, sin2θ + cos2θ = 1 Multiplying both numerator and denominator by (1− sin θ), we have
We know that, sin2A + cos2A = 1 cosec2A - cot2A = 1
= cos2A + sin2A
=1
We know, sin2A + cos2A = 1 sec2A – tan2A = 1
= 1
We know, sin2θ + cos2θ = 1 Multiplying both numerator and denominator by (1+ cos θ), we have
We know that, (a - b)(a + b) = a2 - b2
Here, (1 - cos2θ) = sin2 θ
⟹ cosec θ + cot θ
Hence, L.H.S = R.H.S
Given,
Rationalize with nr and dr with 1 – sin θ
Here, (1 - sin θ)(1 + sin θ) = cos2θ
⟹ (sec θ – tan θ) 2
Hence, L.H.S = R.H.S
Given,
Here, 1 + cot2 θ = cosec2 θ
⟹ cot θ
Hence, L. H. S = R.H.S
tan2θ − sin2θ = tan2θ ∗ sin2θ
Given, L.H.S = tan2θ − sin2θ
⟹ tan2θ *sin2θ
Hence, L.H.S = R.H.S
(cosec θ + sin θ)(cosec θ - sin θ) = cot2θ + cos2θ
Given, L.H.S = (cosec θ + sin θ)(cosec θ - sin θ)
Here, (a + b)(a - b) = a2 - b2 cosec2 θ can be written as 1 + cot2 θ and sin2 θ can be written
as 1 - cos2 θ
⟹ 1 + cot2 θ - (1 - cos2 θ)
⟹ 1 + cot2 θ - 1 + cos2 θ
⟹ cot2 θ + cos2 θ
Hence, L.H.S = R.H.S
(sec θ + cos θ) (sec θ - cos θ) = tan2θ + sin2θ
Given, L.H.S = (sec θ + cos θ)(sec θ - cos θ)
Here, (a + b)(a - b) = a2 - b2 sec2θ can be written as 1 + tan2 θ and cos2 θ can be written as 1 - sin2 θ
⟹ 1 + tan2 θ - (1 - sin2 θ)
⟹ 1 + tan2 θ - 1 + sin2 θ
⟹ tan2θ + sin2 θ
Hence, L.H.S = R.H.S
Sec A(1- sin A) (sec A + tan A) = 1
Given, L.H.S = sec A(1 – sin A)(sec A + tan A)
⟹ 1
Hence, L.H.S = R.H.S
(cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1
Given, L.H.S = sec A(1 – sin A)(sec A + tan A)
Substitute the above values in L.H.S
⟹ 1
Hence, L. H. S = R. H. S
(1 + tan2θ)(1 - sin θ)(1 + sin θ) = 1
Given, L.H.S = (1 + tan2θ)(1 - sin θ)(1 + sin θ)
We know that, Sin2 θ + cos2 θ = 1 and sec2 θ - tan2 θ = 1
So, (1 + tan2θ)(1 - sin θ)(1 + sin θ)
= (1 + tan2θ){(1 - sin θ)(1 + sin θ)}
= (1 + tan2θ)(1 - sin2θ)
Hence, L.H.S = R.H.S
(sin2A ∗ cot2A) + (cos2A; ∗ tan2A) = 1
Given,
= cos2A + sin2A = 1
Hence, L.H.S = R.H.S
1. Give, L.H.S = cot θ - tan θ
Here, Sin2 θ + cos2θ = 1
So, ⟹ cot θ - tan θ
Hence, L.H.S = R.H.S
Given, L. H. S = tan θ – cot θ
We know that,
Sin2θ + cos2θ = 1
tan θ – cot θ
Hence, L. H. S = R. H. S
Given, L.H.S
We know that,
Sin2θ + cos2 θ = 1
= - sin θ + sin θ = 0
Hence, L.H.S = R.H.S
⇒ 2sec2A
LHS = RHS
Hence proved
= 2 sec θ
LHS = RHS
Hence proved
We know that sin2θ + cos2θ = 1
LHS = RHS
Hence prove
LHS = RHS Hence proved
= 1 + cos θ
LHS = RHS Hence proved
= 1 + tan θ + cot θ
LHS = RHS Hence proved
sec6θ = tan6θ + 3 tan2 θ sec2θ + 1
We know that sec2θ − tan2θ = 1
Cubing both sides
(sec2θ − tan2θ)3= 1 sec6θ − tan6θ − 3sec2θ tan2θ(sec2θ − tan2θ) = 1 [Since, a3 − b3 = (a − b)(a2 + ab + b2)]
sec6θ − tan6θ − 3sec2θ tan2θ = 1
⇒ sec6θ = tan6θ + 3sec2θ tan2θ + 1
Hence proved.
cosec6θ = cot6θ + 3cot2θ cosec2θ + 1
We know that cosec2θ − cot2θ = 1
Cubing both sides (cosec2θ − cot2θ)3 = 1
cosec6θ − cot6θ − 3cosec2θ cot2θ (cosec2θ − cot2θ) = 1 [Since, a3 − b3 = (a − b)(a2 + ab + b2)]
cosec6θ − cot6θ − 3cosec2θ cot2θ = 1
⇒ cosec6θ = cot6θ + 3 cosec2θ cot2θ + 1
Hence proved.
We know that sec2θ − tan2θ = 1
Therefore, sec2θ = 1 + tan2θ
LHS = RHS Hence proved
We know that
sin2A + cos2A = 1 sin2A = 1 − cos2A
⇒ sin2A = (1 – cos A)(1 + cos A)
LHS = RHS
Hence proved
Rationalizing the denominator by multiplying and dividing with
Sec A + tan A, we get
LHS = RHS
Hence proved
Multiply both numerator and denominator with (1 – cos A)
LHS = RHS
Hence proved
Considering left hand side (L. H. S), Rationalize the numerator and denominator with
Therefore,
LHS = RHS Hence proved
Considering left hand side (LHS), Rationalize the numerator and denominator.
Therefore, LHS = RHS
Hence proved
Prove that:
Therefore, LHS = RHS
Hence proved
Therefore, LHS = RHS
Hence proved
Considering left hand side (LHS), Rationalize the numerator and denominator.
Therefore, LHS = RHS
Hence proved
Considering left hand side (LHS),
Multiply and divide with (1 + cos θ)
Therefore,
LHS = RHS
Hence proved
Considering left hand side (LHS), = (sec A – tan A)2
Therefore, LHS = RHS Hence proved
Considering left hand side (LHS), Rationalize the numerator and denominator with (1 - cos A)
= (cosec A - cot A)2 = (cot A - cosec)2
Therefore, LHS = RHS Hence proved
Considering left hand side (LHS),
= 2cosec A cot A
Therefore, LHS = RHS
Hence proved
Considering left hand side (LHS),
= cos A + sin A
Therefore, LHS = RHS
Hence proved
Considering left hand side (LHS),
= 2 sec2 A
Therefore, LHS = RHS
Hence proved
Considering left hand side (LHS),
Therefore, LHS = RHS
Hence proved
Considering left hand side (LHS),
Therefore, LHS = RHS
Hence proved
Considering left hand side (LHS),
= cot θ
Therefore, LHS = RHS
Hence, proved.
Dividing the numerator and denominator with cos θ Considering LHS, we get,
Therefore, LHS = RHS
Hence proved
Dividing the numerator and denominator with cos θ, we get,
Therefore, LHS = RHS Hence proved
Considering LHS, we get, dividing the numerator and denominator with sin θ, we get,
= cosec θ + cot θ
Therefore, LHS = RHS
Hence proved
(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ
To prove, (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ
Considering LHS, we get,
= sec θ + cosec θ
Therefore, LHS = RHS Hence proved
To prove,
Considering LHS, we get,
= 2 cosec A
Therefore, LHS = RHS Hence proved
Considering LHS, we get,
= 1 + cosec θ - 1 [∴ (a + b)(a - b) = a2- b2]
= cosec θ [ a = cosec θ, b = 1]
Therefore, LHS = RHS Hence, proved
Therefore, LHS = RHS Hence, proved
Therefore, LHS = RHS. Hence Proved.
sin2A cos2B - cos2A sin2B = sin2A - sin2B
LHS = sin2Acos2B - cos2A sin2B
= sin2A− sin2A sin2 B − sin2 B + sin2A sin2 B
= sin2A − sin2 B
RHS Hence Proved.
= cot A tan B
= RHS Hence Proved.
= tan A tan B
= RHS Hence Proved.
cot2A cosec2B − cot2B cosec2A = cot2A − cot2B
LHS = cot2A cosec2B − cot2 B cosec2A
= cot2A + cot2A cot2B − cot2B − cot2B cot2A
= cot2A - cot2B
RHS Hence Proved.
tan2A sec2B − sec2A tan2B = tan2A − tan2B
LHS = tan2A sec2B − sec2A tan2B
= tan2A(1 + tan2B) − sec2A(tan2A)
= tan2A + tan2A tan2B − tan2B − tan2A tan2B
= tan2A − tan2B
= RHS Hence Proved.
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, Prove that x2 - y2 = a2 - b2.
LHS = x2 - y2 = (a sec θ + b tan θ)2 − (a tan θ + b sec θ)2
= a2sec2θ + b2tan2θ + 2 a b sec θ tan θ − a2 tan2θ − b2sec2θ – 2 ab sec θ tan θ
= a2sec2θ + b2tan2θ − a2tan2θ − b2sec2θ
= a2sec2θ − b2sec2θ + b2tan2θ − a2tan2θ
= sec2θ (a2 − b2) + tan2θ (b2 − a2)
= sec2θ (a2 − b2) − tan2θ (a2 − b2)
= (sec2θ − tan2θ) (a2 − b2) = a2 - b2
RHS Hence Proved.
If 3 sin θ + 5 cos θ = 5,
Prove that 5 sin θ – 3 cos θ = ± 3.
Given 3 sin θ + 5 cos θ = 5
3 sin θ = 5 – 5 cos θ
3 sin θ = 5(1 – cos θ)
3+ 3 cos θ = 5sin θ
5 sin θ - 3 cos θ = ±3
= RHS Hence Proved.
5 sin θ – 3 cos θ = ± 3.
If cosec θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1.
LHS = m n = (cosec θ + cot θ) (cosec θ – cot θ)
= cosec2θ − cot2θ
= 1
= RHS Hence Proved.
If Tn = sinn θ + cosn θ, Prove that
= sin2θ cos2θ
LHS = RHS
Hence proved.
(tan θ + sec θ)2 + (tan θ – sec θ)2 = tan2θ + sec2θ + 2 tan θ sec θ + tan2 θ + sec2θ – 2 tan θ sec θ
= 2 tan 2θ + 2 sec 2θ
= 2[tan2θ + sec2θ]
LHS = RHS Hence proved.
LHS = RHS Hence proved.
LHS = RHS Hence proved.
LHS = RHS
Hence proved.
(sec A + tan A - 1) (sec A - tan A + 1) = 2 tan A
= (sec A + tan A − {sec2A - tan2A}) [sec A – tan A + (sec2A − tan2A)]
= (sec A + tan A − (sec A + tan A) (sec A – tan A)) [sec A – tan A + (sec A – tan A)(sec A + tan A)]
= (sec A + tan A) (1 − (sec A – tan A)) (sec A – tan A) (1 + (sec A + tan A))
= (sec2A − tan2A) (1 – sec A + tan A) (1 + sec A + tan A)
=2 tan A
LHS = RHS Hence proved.
(1 + cot A - cosec A)(1 + tan A + sec A) = 2
LHS = (1 + cot A - cosec A) (1 + tan A + sec A)
LHS = RHS Hence proved.
(cosec θ – sec θ) (cot θ – tan θ) = (cosec θ + sec θ) (sec θ cosec θ − 2)
LHS = (cosec θ – sec θ) (cot θ – tan θ)
RHS = (cosec θ + sec θ) (sec θ cosec θ - 2)
LHS = RHS Hence proved.
= cosec A - sec A
= RHS = LHS Hence proved.
= LHS = RHS Hence proved.
= sin A × cos3 A +cos A × sin3A
= sin A cos A (cos2A + sin2A)
= sin A cos A
LHS = RHS Hence proved.
sec4A (1 − sin4A) – 2 tan2A = 1
Given, L.H.S = sec4A (1 - sin4A) - 2tan2A
= sec4A − sec4A × sin4A − 2tan4A
= sec4A - tan4A – 2 tan4A
= (sec2A)2 - tan4A – 2 tan4A
= (1+ tan2A)2 − tan4A − 2tan4A
= 1 + tan4A + 2tan2A − tan4A − 2tan4A
= 1
Hence, L.H.S = R.H.S
Here, sin2A + cos2A = 1
Solving, RHS ⟹
Multiplying Nr. And Dr. with (1 + Sin A)
Hence, LHS = RHS
Given, L.H.S
= (1 + cot A + tan A) (sin A – cos A)
⟹ sin A – cos A + cot A sin A – cot A cos A + sin A tan A – tan A cos A
⟹ sin A tan A – cos A cot A
Hence, L.H.S = R.H.S
If cosec θ – sin θ = a3, sec θ – cos θ = b3, Prove that a2b2 (a2+ b2) = 1
Given, cosec θ – sin θ = a3
Here cos2θ A + sin2 A = 1
Squaring on both sides
Squaring on both sides
= 1
Hence, L.H.S = R.H.S
If a cos3 θ + 3 a cos θ sin2θ = m, a sin3θ + 3 a cos2θ sin θ = n,
Substitute the values of m and n in the above equation
⟹ a cos3θ + 3 a cos θ sin2 θ + a sin3 θ + 3 a cos2θ sin2/3 θ + a cos3 θ + 3 a cos θ sin2 θ – a sin3 θ – 3 a cos2 θ sin2/3 θ
⟹ (a)2/3 cos3θ + 3 cos θ sin2 θ + sin3θ + 3 cos2 θ sin2/3θ + (a)2/3 cos3 θ + 3 cos θ sin 2θ - sin3 θ – 3 cos2 θ sin2/3θ
⟹ (a)2/3((cos θ + sin θ)3)2/3 + (a)2/3 ((cos θ – sin θ)3)2/3
⟹ (a)2/3[(cos θ + sin θ)2] + (a)2/3[(cos θ – sin θ)2]
⟹ (a)2/3((cos2θ + sin2θ + 2 sin θ cos θ)) + (a)2/3((cos2θ + sin2θ – 2 sin θ cos θ))
⟹ (a)2/3[ 1 + 2 sin θ cos θ] + (a)2/3[1 – 2 sin θ cos θ]
⟹ (a)2/3[1 + 2 sin θ cos θ] + 1 – 2 sin θ cos θ]
⟹ (a)2/3(1 + 1)
⟹ 2(a)2/3
Hence, L.H.S = R.H.S
If x = a cos3θ, y = b sin3θ,
x = a cos3 θ: y = b sin3 θ
= 1
Hence proved.
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, Prove that a2 + b2 = m2 + n2
R.H.S = m2 + n2 = m2 + n2
= (a cos θ + b sin θ)2 + (a sin θ – b cos θ)2
= a2 cos2 θ + b2 sin2θ + 2ab sin θ cos θ + a2 + sin2θ + b2 cos2θ – 2ab sin θ cos θ
= a2 cos2θ + b2 cos2θ + b2 sin2θ + a2 sin2θ
= a2(sin2 θ + cos2 θ) + b2(sin2 θ + cos2 θ)
= a2 + b2 (∵ sin2 θ + cos2θ = 1)
If cos A + cos2A = 1, Prove that sin2A + sin4A = 1
Given - cos A + cos2 A = 1
We have to prove sin2 A + sin4 A = 1
Now, cos A + cos2 A = 1
cos A = 1 - cos2 A
cos A = sin2 A
sin2 A = cos A
Therefore,
We have sin2 A + sin4 A = cos A + (cos A)2
= cos A + cos2 A = 1
Hence proved.
If cos θ + cos2 θ = 1, Prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
cos θ + cos2 θ = 1
cos θ = 1 − cos2 θ
cos θ = sin2 θ (i)
Now, sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2
= (sin4 θ)3 + 3 sin4 θ. sin2 θ (sin4 θ + sin2 θ) + (sin2 θ)3 + 2(sin2 θ)2 + 2 sin2 θ − 2
Using (a + b)3 = a3 + b3 + 3 ab(a + b) and also from(i) cos θ = sin2 θ
(sin4 θ + sin2 θ)3 + 2cos2 θ + 2 cos θ − 2
((sin2 θ)2 + sin2 θ)3 + 2 cos2 θ + 2 cos θ – 2
(cos2 θ + sin2 θ)3 + 2 cos2 θ + 2 cos θ − 2
1 + 2(cos2 θ + sin2 θ) − 2
= 1 + 2(1) −2
= 1
L.H.S = R.H.S Hence proved.
Given that: (1 + cos α)(1 + cos β)(1 + cos γ) = (1 – cos α)(1 – cos β)(1 – cos γ). Show that one of the values of each member of this equality is sin α sin β sin γ.
We know that
Multiply (i) with sin α sin β sin γ and divide it with same we get
Hence sin α sin β sin γ is the member of equality.
If sin θ + cos θ = x,
sin θ + cos θ = x
Squaring on both sides
(sin θ + cos θ)2 = x2
⇒ sin2θ + cos2θ + 2 sin θ cos θ = x2
We know that sin2 θ + cos2 θ = 1
Cubing on both sides (sin2 θ + cos2 θ)3 = 13
sin6 θ + cos6 θ + 3 sin2 θ cos2 θ (sin2 θ + cos2 θ) = 1
⇒ sin6 θ + cos6 θ = 1 – 3 sin2 θ cos2θ
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan ϕ,
Given, x = a sec θ cos ϕ, y = b sec θ sin ϕ, z = c tan ϕ squaring x, y, z on the sides
x2 = a2sec2θ cos2 ϕ x2/a2 = sec2θ cos2 ϕ --- 1
y2 = b2 sec2θ sin2 ϕ y2/b2 = sec2θ sin2 ϕ --- 2
z2 = c2 tan2 ϕ z2/c2 = tan2 ϕ --- 3
Substitute equation 1, 2, 3 in x2/a2 + y2/b2 − z2/c2
⟹ x2/a2 + y2/b2 − z2/c2
⟹ Sec2 θ cos2 ϕ + sec2 θ sin2 ϕ - tan2 ϕ
⟹ sec2θ(cos2 ϕ + sin2 ϕ) - tan2 ϕ
We know that,
cos2 ϕ + sin2 ϕ = 1
⟹ sec2θ (1) - tan2 ϕ and sec2θ − tan2θ = 1
⟹ 1
Hence, L.H.S= R.H.S
If sin θ + 2 cos θ. Prove that 2 sin θ – cos θ = 2
Given, sin θ + 2 cos θ = 1
Squaring on both sides
⟹ (sin θ + 2 cos θ)2 = 12
⟹ sin2θ + 4 cos2θ + 4 sin θ cos θ = 1
⟹ 4 cos2 θ + 4 sin θ cos θ = 1 - sin2θ
Here, 1 - sin2θ = cos2θ
⟹ 4 cos2θ + 4 sin θ cos θ - cos2θ = 0
⟹ 3 cos2θ + 4 sin θ cos θ = 0 ---- 1
We have,
2 sin θ – cos θ = 2
Squaring L.H.S
(2 sin θ – cos θ)2
= 4 sin2θ + cos2θ – 4 sin θ cos θ
Here, 4 sin θ cos θ = 3 cos2θ
= 4 sin2θ + cos2θ + 3 cos2θ
= 4 sin2 θ + 4 cos2 θ
= 4(sin2θ + cos2θ)
= 4(1)
= 4 (2 sin θ – cos θ)2 = 4
⟹ 2sin θ – cos θ
= 2
Hence proved