Trigonometric Functions

To prove:

As equation on RHS is a simplified expression, so we must opt Left side equation and simplify it further so that we can get

LHS = RHS.

And thus we will be able to prove it.

∵ LHS =

The expression gives us hint that we ca use the identity:

sin²A + cos²A = 1 to simplify the expression but to use this we need to multiply numerator and denominator with

sin A + (1 – cos A).

∴ LHS =

Hence, it is proved that:

As we have to prove:

Given,

y =

As we need to get (1 – cos α + sin α) in numerator, so we need to bring this term in numerator.

∴ Multiplying numerator and denominator by

(1 – cos α + sin α),we get -

using (a + b) (a-b) = a² – b², we get:

∵ 1 – cos²α = sin² α

Thus,

We know that,

a) is correct since Sin θ ∈ [-1,1]

b) cos θ = 1 is correct since Cos θ ∈ [-1,1]

c)

⇒cos θ=2 is incorrect since Cos θ ∈ [-1,1]

d) tan θ = 20 is correct since tan θ ∈ R.

e)

Given,

m sin θ = n sin (θ + 2α)

To prove: tan (θ + α)cot α =

∵ m sin θ = n sin (θ + 2α)

⇒

Applying componendo-dividendo rule –

⇒

By transformation formula of T-ratios we know that –

sin A + sin B =

and sin A – sin B =

∴ On applying formula, we get –

⇒

⇒ {∵ tan x = (sin x)/(cos x)}

∴ tan (θ + α) cot α =

Given that,

cos(α + β) = 4/5 …(1)

we know that: sin x = √(1 – cos²x)

∴ sin (α + β) = √(1 – cos²(α + β))

⇒ sin (α + β) = …(2)

Also,

sin(α - β) = 5/13 {given} …(3)

we know that: cos x = √(1 – sin²x)

∴ cos (α - β) = √(1 – sin²(α - β))

⇒ cos (α - β) = …(4)

∵ tan 2α = tan (α + β + α – β)

We know that:

∴ tan 2α =

⇒ tan 2α =

Using equation 1,2,3 and 4 we have –

Hence, tan 2α = 56/33

Given: tan x = b/a

Let, y =

To prove:

As equation on RHS is a simplified expression, so we must opt Left side equation and simplify it further so that we can get

LHS = RHS.

And thus we will be able to prove it.

∵ LHS =

By seeing the expression we can think that the problem can be solved using transformation formula:

By transformation formula, we have:

2 cos A cos B = cos(A + B) + cos (A – B)

-2 sin A sin B = cos(A + B) - cos (A – B)

Or cos A – cos B =

But as LHS expression does not contain ‘2’ in its term. So we multiply and divide the expression by 2.

∴ LHS =

Applying transformation formula, we have –

LHS =

⇒ LHS =

⇒ LHS = {∵ cos (-x) = cos x}

⇒ LHS =

Again, applying the transformation formula:

⇒ LHS =

⇒ LHS =

∴ LHS = sin 4θ sin = RHS

Hence:

Given,

a cos θ + b sin θ = m …(1)

a sin θ – b cos θ = n …(2)

Squaring and adding equation 1 and 2, we get –

(a cos θ + b sin θ)² + (a sin θ – b cos θ)² = m² + n²

⇒ a²cos²θ + b²sin²θ + 2ab sin θ cos θ + a²sin²θ + b²cos²θ - 2ab sin θ cos θ = m² + n²

⇒ a²cos²θ + b²sin²θ + a²sin²θ + b²cos²θ = m² + n²

⇒ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ) = m² + n²

Using: sin²θ + cos²θ = 1, we get –

⇒ a² + b² = m² + n²

Let, θ = 45°

As we need to find: tan 22°30’ = tan (θ/2)

∵

If somehow, we manage to get sin θ and cos θ in numerator and denominator, our problem will be solved as we know that –

sin θ = cos θ = 1/√2 (for θ = 45°)

As,

Multiplying in numerator and denominator, we get –

⇒

By applying formula of T-ratios of multiple angles-

sin 2x = 2sin x cos x

cos 2x = 2cos²x – 1 or 1 + cos 2x = 2cos²x

∴

⇒ tan 22°30’ =

By rationalizing the term, we get-

⇒ tan 22°30’ =

∴ tan 22°30’ = √2 – 1

∵ sin 4A = sin (2A + 2A)

As we know that-

sin(A + B) = sin A cos B + cos A sin B

∴ sin 4A = sin 2A cos 2A + cos 2A sin 2A

⇒ sin 4A = 2 sin 2A cos 2A

From T-ratios of multiple angle, we know that

sin 2A = 2 sin A cos A and cos 2A = cos²A – sin²A

⇒ sin 4A = 2(2 sin A cos A)(cos²A – sin²A)

⇒ sin 4A = 4 sin A cos³A – 4 cos A sin³A

Hence: sin 4A = 4 sin A cos³A – 4 cos A sin³A

Given,

tan θ + sin θ = m …(1)

tan θ – sin θ = n …(2)

As we have to prove: m² – n² = 4 sin θ tan θ

∴ if we find out the expression of sin θ and tan θ in terms of m and n, we can get the desired expression to be proved.

∴ Adding equation 1 and 2 to get the value of tan θ.

2 tan θ = m + n …(3)

Similarly, on subtracting equation 2 from 1, we get-

2sin θ = m – n …(4)

Multiplying equation 3 and 4 –

2sin θ (2tan θ) = (m + n)(m – n)

⇒ 4 sin θ tan θ = m² – n²

Hence,

m² – n² = 4 sin θ tan θ

To prove:

∵ tan 2A = tan (A + B + A – B)

We know that:

∴ tan 2A =

⇒ tan 2A = {according to values given in question}

Hence, tan 2A =

To Prove: cos 2α + cos 2β = –2cos (α + β)

Given,

cosα + cosβ = 0 = sinα + sinβ …(1)

∵ LHS = cos 2α + cos 2β

We know that: cos 2x = cos²x – sin²x

∴ LHS = cos²α – sin²α + (cos²β – sin²β)

⇒ LHS = cos²α + cos²β – (sin²α + sin²β)

∵ a² + b² = (a+b)² – 2ab

⇒ LHS = (cosα + cosβ)² – 2cosα cosβ –(sinα + sinβ)² +2sinα sinβ

⇒ LHS = 0 - 2cosα cosβ -0 + 2sinα sinβ {using equation 1}

⇒ LHS = -2(cosα cosβ – sinα sinβ)

∵ cos (α + β) = cosα cosβ – sinα sinβ

∴ LHS = -2 cos (α + β) = RHS

Hence, cos 2α + cos 2β = –2cos (α + β)

To prove:

Given,

∵ sin(A+B) = sin A cos B + cos A sin B

∴

⇒

Applying componendo-dividendo rule, we get –

⇒

⇒

⇒

⇒ {∵ tan A = (sinA)/(cosA)}

⇒

Given,

To prove: sinα + cosα = √2 cos θ

∵

⇒

⇒ {∵ tan A = (sinA)/(cosA)}

⇒ {∵ tan π/4 = 1}

We know that: tan(x-y) =

∴

⇒ θ = α - π/4

⇒ α = θ + π/4 …(1)

As we have to prove - sinα + cosα = √2 cos θ

∵ LHS = sinα + cosα

⇒ LHS = sin(θ + π/4) + cos(θ + π/4) {using equation 1}

∵ sin(x + y) = sin x cos y + cos x sin y

And, cos(x + y) = cos x cos y – sin x sin y

∴ LHS = sin θ cos(π/4) + sin(π/4)cos θ + cos θ cos(π/4) - sin(π/4)sin θ

∵ sin(π/4)=cos(π/4) = 1/√2

⇒ LHS =

⇒ LHS =

⇒ LHS = √2 cos θ = RHS

Hence, sinα + cosα = √2 cos θ

Given,

sin θ + cos θ = 1

We need to solve the above equation.

If we can convert this to a single trigonometric ratio, we can easily give its solution by using the formula.

∵ sin θ + cos θ = 1

⇒

⇒ {∵ sin(π/4)=cos(π/4) = 1/√2}

We know that: sin(A+B) = sinAcosB + cosAsinB

⇒

⇒

Formula to be used: If sin θ = sinα ⇒ θ = nπ + (-1)ⁿα

∴ θ + π/4 = nπ + (-1)ⁿ(π/4)

⇒ θ = nπ + (π/4)((-1)ⁿ – 1)

As we need to find the most general solution for two different trigonometric equations.

Tan θ = -1 …(1)

and cos θ = 1/√2 …(2)

Most general value of θ is the common solution of both equation 1 and 2.

Let S₁ represents the solution set of equation 1 and S₂ be the solution set equation 2.

Let S represents the set of most general value of θ

∴ S = S₁ ∩ S₂

We know that, solution of tan x = tan α is given by –

x = nπ + α ∀ n ∈ Z

and solution of cos x = cos α is given by

x = 2mπ ± α ∀ m ∈ Z

From equation 1,we have –

tan θ = -1

⇒ tan θ = tan (-π/4)

∴ θ = nπ + (-π/4) = (nπ - π/4) ∀ n ∈ Z …(3)

From equation 2 we have -

cos θ = 1/√2

⇒ cos θ = cos π/4

⇒ θ = 2mπ ± π/4 ∀ m ∈ Z …(4)

From equation 3 we can infer that, solution lies either in 2^nd quadrant (when n is odd) and 4^th quadrant (when n is even)

From equation 3 we can infer that, solution lies either in 1^st or 2^nd quadrant irrespective of value of m.

∴ region of common solution is 4^th quadrant and n is even in that case.

∴ common solution is given by-

θ = 2mπ - π/4 ∀ m ∈ Z.

The above solution is the most general solution for the given equations.

Given,

⇒

⇒ {∵ sin²θ + cos²θ = 1}

⇒ 1 = 2 cosec θ sin θ cos θ

⇒ 1 = 2 cos θ {∵ sin θ cosec θ = 1}

⇒ cos θ = 1/2 = cos(π/3)

∵ solution of cos x = cos α is given by

x = 2mπ ± α ∀ m ∈ Z

⇒ θ = 2nπ ± π/3, n ∈ Z

Given,

2sin²θ = 3cos θ

We know that – sin²θ = 1 – cos²θ

∴ 2(1 – cos²θ) = 3cos θ

⇒ 2 – 2cos²θ = 3cos θ

⇒ 2cos²θ + 3cos θ - 2 = 0

⇒ 2cos²θ + 4cos θ - cos θ - 2 = 0

⇒ 2cos θ (cos θ+ 2) +1 (cos θ + 2) = 0

⇒ (2cos θ + 1)(cos θ + 2) = 0

∵ cos θ ∈ [-1,1] , for any value θ.

So, cos θ ≠ - 2

∴ 2 cos θ - 1 = 0

⇒ cos θ = 1/2

Given,

sec x cos 5x = -1

⇒ cos 5x = -1/sec x

⇒ cos 5x + cos x = 0 {∵ sec x = 1/cos x}

By transformation formula of T-ratios we know that –

cos A + cos B =

⇒

⇒ 2 cos 3x cos 2x = 0

⇒ cos 3x = 0 or cos 2x = 0

∵ 0 < x ≤ π/2

∴ 0< 2x ≤ π or 0< 3x ≤ 3π/2

∴ 2x = π/2

⇒ x = π/4

3x = π/2

⇒ x = π/6

Or 3x = 3π/2

⇒ x = π/2

Hence, x = π/6, π/4.

Given,

sin (θ + α) = a and sin(θ + β) = b

To prove: cos 2(α – β) – 4ab cos (α – β) = 1 – 2a² – 2b²

As, LHS = cos 2(α – β) – 4ab cos (α – β)

Using: cos 2x = 2cos²x – 1

⇒ LHS = 2cos²(α – β) - 1 – 4ab cos(α – β)

⇒ LHS = 2cos (α – β) {cos (α – β) – 2ab} - 1

It involves cos (α – β) terms so we need to calculate it first.

∵ cos (α – β) = cos {(θ + α) – (θ + β)}

cos (A – B) = cos A cos B + sin A sin B

⇒ cos (α – β) = cos(θ + α)cos(θ + β) + sin(θ + α)sin(θ + β)

∵ sin(θ + α) = a

⇒ cos(θ + α) = √(1 – sin²(θ + α) = √(1 – a²)

Similarly, cos(θ + β) = √(1 – b²)

∴ cos(α – β) = √(1-a²)√(1-b²) + ab

∴ LHS = 2{ab + √(1 – a²)(1 – b²)}{ab + √(1 – a²)(1 – b²) -2ab} - 1

⇒ LHS = 2{√(1 – a²)(1 – b²) + ab}{√(1 – a²)(1 – b²) - ab}-1

Using: (x + y)(x – y) = x² – y²

⇒ LHS = 2{(1-a²)(1-b²) – a²b²} – 1

⇒ LHS = 2{1 – a² – b² + a²b²} – 1

⇒ LHS = 2 – 2a² – 2b² – 1

⇒ LHS = 1 – 2a² – 2b² = RHS

Thus,

cos 2(α – β) – 4ab cos (α – β) = 1 – 2a² – 2b²

Given,

cos (θ + ϕ) = m cos (θ – ϕ)

To prove:

∵ cos (θ + ϕ) = m cos (θ – ϕ)

⇒

Applying componendo – dividend, we get

⇒

From transformation formula, we know that –

cos(A+B) + cos(A – B) = 2cosAcosB

cos(A – B) – cos(A + B) = 2sinA sinB

⇒

⇒ {∵ (cos θ)/(sin θ) = cot θ }

⇒

⇒

Let, y =

We know that-

sin(3π/2 – α) = -cos α

sin(3π + α) = -sin α

sin(π/2 + α) = cos α

sin(5π – α) = sin α

∴ y =

⇒ y = 3 [cos⁴α + sin⁴α] – 2[sin⁶α + cos⁶α]

⇒ y = 3[(sin²α + cos²α)² – 2sin²α cos²α] – 2[(sin²α)³ + (cos²α)³]

∵ sin²α + cos²α = 1

⇒ y = 3[1 – 2sin²α cos²α] – 2[(sin²α + cos²α)( cos⁴α + sin⁴α- sin²α cos²α)] {∵ a³+b³ = (a+b)(a² – ab + b²)}

⇒ y = 3[1 – 2sin²α cos²α] – 2[cos⁴α + sin⁴α- sin²α cos²α]

⇒ y = 3[1 – 2sin²α cos²α] – 2[(sin²α + cos²α)² – 2sin²α cos²α - sin²α cos²α]

⇒ y = 3[1 – 2sin²α cos²α] – 2[1 – 3sin²α cos²α]

⇒ y = 3 – 6sin²α cos²α – 2 + 6 sin²α cos²α

⇒ y = 1

Given,

a cos2θ + b sin 2θ = c and α and β are the roots of the equation.

Using the formula of multiple angles, we know that –

and

∴

⇒ a(1 – tan²θ) + 2b tan θ - c(1 + tan²θ) = 0

⇒ (-c – a)tan²θ + 2b tan θ - c + a = 0 …(1)

Clearly it is a quadratic equation in tan θ and as α and β are its solutions.

∴ tan α and tan β are the roots of this quadratic equation.

We know that sum of roots of a quadratic equation:

ax² + bx + c = 0 is given by (-b/a)

∴ tan α + tan β =

Hence,

Given,

x = sec ϕ – tan ϕ and y = cosec ϕ + cot ϕ

To prove: xy + x – y + 1 = 0

∵ LHS = xy + x – y + 1

{∵ sin²θ + cos²θ = 1}

Thus, LHS = xy + x – y + 1 = 0

Given,

cos θ = 8/17

∴ sin θ = ±√(1 – cos²θ)

But θ lies in first quadrant, so only positive sign is considered.

⇒ sin θ = √( 1 – 64/289) = 15/17

Let, y = cos(30° + θ) + cos (45° – θ) + cos (120° – θ)

We know that: cos(x + y) = cos x cos y – sin x sin y

∴ y = cos30° cos θ – sin30° sin θ + cos45° cos θ + sin45°sin θ +cos120° cos θ + sin120° sin θ

Putting values of cos30° ,sin30° ,cos 120° ,sin120° and cos 45°

Let, y =

=

∵ cos (π – x) = - cos x

Given,

5cos²θ + 7sin²θ – 6 = 0

We know that : sin²θ = 1 – cos²θ

∴ 5cos²θ + 7(1 – cos²θ) – 6 = 0

⇒ 5cos²θ + 7 – 7cos²θ – 6 = 0

⇒ -2cos²θ + 1 = 0

⇒ cos²θ = 1/2

∴ cos θ = ±1√2

∴ cos θ = cos π/4 or cos θ = cos 3π/4

∵ solution of cos x = cos α is given by

x = 2mπ ± α ∀ m ∈ Z

⇒ θ = nπ ± π/4, n ∈ Z

Given,

sin x – 3sin2x + sin3x = cos x – 3cos2x + cos3x

To solve the problem we need to apply transformation formula, but we need to group sin x and sin 3x together and similarly cos x and cos 3x in RHS side.

∴ sin x + sin3x – 3sin2x = cos x + cos3x – 3cos2x

Transformation formula:

cos A + cos B =

sin A + sin B =

⇒

⇒ 2sin 2x cos x – 3sin 2x = 2cos 2x cos x – 3cos 2x

⇒ 2sin 2x cos x – 3sin 2x - 2cos 2x cos x + 3cos 2x = 0

⇒ 2cos x (sin 2x – cos 2x) -3(sin 2x – cos 2x) = 0

⇒ (sin 2x – cos 2x)(2cos x – 3) = 0

⇒ cos x = 3/2 or sin 2x = cos 2x

As cos x ∈ [-1,1]

∴ no value of x exists for which cos x = 3/2

∴ sin 2x = cos 2x

⇒ tan 2x = 1 = tan π/4

We know solution of tan x = tan α is given by –

x= nπ + α , n ∈ Z

∴ 2x = nπ + (π/4)

⇒

Given equation is:

We need to solve the above equation.

If we can convert this to a single trigonometric ratio, we can easily give its solution by using the formula.

For this,

Let, r sinα = √3 – 1 and r cosα = √3 + 1

∴

And,

Hence, equation can be rewritten as-

∴ r(sinα cos θ + r cosα sin θ) = 2

⇒ 2√2 sin (θ + α) = 2

⇒ sin(θ + α) = 1/√2 = sin π/4

We know that: General solution of trigonometric equation

sin x = sin α is given as –

x = nπ + (-1)ⁿα , n ∈ Z

∴ θ + α = nπ + (-1)ⁿ(π/4)

⇒ θ = {nπ + (-1)ⁿ(π/4) – α} ,where tan α = and n ∈ Z

Given that, sin θ + cosec θ = 2

Squaring both sides, we get

⇒ (sin θ + cosec θ)² = 2²

⇒ sin²θ + cosec²θ + 2 sin θ cosec θ = 4

⇒ sin²θ + cosec²θ + 2 = 4

⇒ sin²θ + cosec²θ = 2

We have, f(x) = cos²x + sec²x

We know that, A.M ≥ G.M.

⇒ cos0²x + sec²x ≥ 2

⇒ f(x) ≥ 2

We have,

We know that,

tan 1° tan 2° tan 3°… tan 89°

= tan 1° tan 2° … tan 45° tan (90-44°) tan(90-43°)…tan (90-1°)

= tan 1°tan 2° … tan 45°cot 44°cot 43°…cot 1° [∵tan (90-θ)=cot θ]

= tan 1° cot 1° tan 2° cot 2°…tan45°… tan 89° cot 89°

=1.1….1 = 1

Let θ = 15° ⇒ 2θ = 30°

We know that,

Since cos90° =0

⇒ cos 1° cos 2° cos 3°… cos90°… cos 179° = 0

Given that, tan θ = 3 and θ lies in third quadrant

We know that,

Cosec²θ = 1+cot²θ

Given that, tan 75°– cot 75°

= -2cot150°

= -2 cot (180°-30°)

= 2cot30°

=2√3

We know that,

57° lies between 0 and 90 degrees and since in first quadrant sin θ increases when θ increases.

⇒ sin 1°< sin 1

Given that,

Now,

Let y = 3 cos x + 4 sin x + 8

⇒ y-8 = 3 cos x + 4 sin x

We know that, minimum value of A cos θ +Bsin θ = -√(A²+B²)

⇒ y-8 = - √(3²+4²) = -√25 = -5

⇒ y = -5+8

= 3

tan 3A = tan(2A+A)

⇒ tan 3A(1-tan 2A.tan A) = tan 2A + tan A

⇒ tan 3A-tan 3A.tan 2A.tan A = tan 2A + tan A

⇒ tan 3A- tan 2A - tan A = tan 3A.tan 2A.tan A

We have, sin (45° + θ) – cos (45°–θ)

= sin (45° + θ) - sin (90° -(45° - θ))

= sin (45° + θ) - sin (45° + θ)

= 0.

We have,

Given that, cos 2θ cos 2ϕ + sin²(θ–ϕ) – sin²(θ + ϕ)

= cos 2θ cos 2ϕ + sin(θ–ϕ+ θ + ϕ)sin(θ–ϕ- θ – ϕ)

[∵sin²A – sin²B = sin (A + B) sin (A – B)]

= cos 2θ cos 2ϕ + sin 2θ. sin(- 2ϕ)

= cos 2θ cos 2ϕ - sin 2θ. Sin 2ϕ [∵ sin(-θ) = -sin θ]

= cos 2(θ + ϕ) [∵ cos x cos y – sin x sin y = cos(x + y)]

We have, cos 12° + cos 84° + cos 156° + cos 132°

= (cos 132°+ cos 12°) + (cos 156° + cos 84°)

= 2 cos72°.cos60° + 2 cos120°.cos36°

= cos72° - cos36°

= cos(90° -18°) – cos36°

= sin 18° - cos36°

Given that,

We know that.

Now,

=3

We have,

=-sin18° sin(90° - 36°)

=-sin18° cos36°

We have, sin 50°– sin 70° + sin 10°

=2cos60° (-sin10°) + sin10°

=-sin10° + sin10°

= 0

Given that, sin θ + cos θ = 1

Squaring both sides, we get

⇒ (sin θ + cos θ)² = 1

⇒ sin²θ + cos²θ + 2sin θ cos θ = 1

⇒ 1 + sin2θ = 1

⇒ sin2θ = 1-1

= 0

We have,

⇒ tanα + tanβ = 1- tanα. tanβ

⇒ tanα + tanβ + tanα.tanβ = 1

On adding 1 both sides, we get

⇒1+ tanα + tanβ + tanα. tanβ = 1+1

⇒1(1+ tanα) + tanβ (1 + tanα) = 2

⇒ (1+ tanα) (1 + tanβ) = 2

Given that, and θ lies in third quadrant.

since θ lies in third quadrant

We have, tan x + sec x = 2 cos x

⇒ 1+sin x = 2cos²x

⇒ 2cos²x-sin x-1 = 0

⇒ 2(1-sin²x)-sin x-1 = 0

⇒ 2-2sin²x-sin x-1 = 0

⇒ 2sin²x+sin x-1 = 0

Since, the equation is a quadratic equation in sin x, it will have 2 solutions.

Given that,

We have, 3 tan A + 4 = 0, A lies in second quadrant.

since A lie in second quadrant.

Now, 2 cot A– 5 cos A + sinA

Given that, cos² 48°– sin² 12° = cos (48° + 12°) cos (48° – 12°)

[∵cos²A – sin²B = cos (A + B) cos (A – B)]

= cos60° cos36°

Given that,

…. (i)

Now

….. (ii)

…. (iii)

…… (iv)

Hence, from (i) and (iii)

Cos2α = sin4β

Given that,

Now,

=b

Given that,

⇒ x²+1 = x cos θ

⇒ x²- cos θx +1 = 0

We know that, b²-4ac ≥ 0

⇒ (-cos θ)²-4×1×1 ≥ 0

⇒ cos²θ -4 ≥ 0

⇒ cos²θ ≥ 4

⇒ cos θ ≥ ± 2

But -1 ≤ cos θ ≤ 1

So, no value of θ is possible

=1

Given that,

⇒ k = sin 10° sin 50° sin 70°

= sin 10° sin (90° - 40°). sin (90° - 20°)

= sin 10° cos 40° cos 20°

[∵ 2 cos x cos y = cos(x + y)+cos(x-y)]

[∵ 2 sin x cos y = sin (x + y)+sin(x-y)]

Given that,

Now,

= tan B

⇒ tan 2A = tan B

Given that, sin x + cos x = a

Squaring both sides, we get

(sin x + cos x)² = a²

⇒ sin²x + cos²x + 2sin x cos x = a²

⇒ 1 + 2sin x cos x = a²

(i) sin⁶ x + cos⁶x = (sin²x)³ + (cos²x)³

= (sin²x+ cos²x)³ -3 sin²x cos²x(sin²x+ cos²x)

(ii) |sin x – cos x| =

|sin x – cos x|² = sin²x + cos²x - 2sin x cos x

=1-(a² -1)

= 1-a²+1 = 2- a²

⇒ |sin x – cos x| =

Given that, a ΔABC with ∠C = 90° and an equation whose roots are tan A and tan B

⇒ x² – (tan A + tan B)x + tan A. tan B = 0 (i)

Now, A+B = 90° [∵ ∠C = 90°]

⇒ tan(A+B) = tan90°

⇒ 1-tan A tan B = 0

⇒ tan A tan B = 1 (ii)

Now,

Substituting the values from (ii) and (iii) in equation (i), we get

We have, 3(sin x – cos x)⁴ + 6 (sin x + cos x)² + 4(sin⁶ x + cos⁶ x)

=3(sin²x+cos²x-2sin x cos x)²+6(sin²x+cos²x+2sin x cos x) +4[(sin²x)³ + (cos²x)³]

=3(1-2sin x cos x)²+6(1+2sin x cos x) +4[(sin²x+ cos²x)³ -3 sin²x cos²x(sin²x+ cos²x)]

=3(1+4 sin²x cos²x-4sin x cos x) +6+12sin x cos x+4-12sin²x cos²x

=3+12 sin²x cos²x-12 sin x cos x+6+12sin x cos x+4-12sin²x cos²x

=3+6+4

= 13

We have,

Let

∴ f(x) = -3cosy

We know that, -1 ≤ cos y ≤ 1

⇒ -3 ≤ 3cosy ≤ 3

⇒ 3 ≥ -3cosy ≥ -3

⇒ -3 ≤ -3cosy ≤ 3

Hence,

We have, y = √3 sin x + cos x

The maximum distance of a point on the graph of the function y = √3 sin x+cos x from x-axis is

=2

True

Explanation:

We have,

⇒ tan 2A = tan B

False

Explanation:

We know that, maximum value of sin A is 1.

Given that, sin A + sin 2A + sin 3A = 3

It is possible when sin A = sin 2A = sin 3A=1

But sin2A and sin3A is not equal to 1.

Hence, statement is false.

False

Explanation:

Let, sin 10° > cos 10°

⇒ sin 10° > cos (90° - 80°)

⇒ sin 10° > sin80°

Which is not possible since value of sin increases with increase in θ.

Hence, statement is false.

True

Explanation:

We have,

=cos24° cos48° cos96° cos192°

=RHS

False

Explanation:

We have, sin⁴θ – 2sin²θ – 1 = 0

Now,

⇒ sin²θ = 1±√2

⇒ sin²θ = 1+√2 or 1-√2

We know that, -1≤ sin θ ≤ 1

⇒ sin²θ ≤ 1

but sin²θ = 1+√2 or 1-√2

Which is not possible.

Hence, the above statement is false.

True

Explanation:

We have, cosec x = 1 + cot x

⇒ sin x+cos x = 1

=2nπ

Hence, statement is true.

True

Explanation:

We have, tan θ + tan 2θ + tan θ tan 2θ =

_⇒ tan θ + tan 2θ = -√3 tan θ tan2θ +√3

_⇒ tan θ + tan 2θ = √3(1- tan θ tan2θ)

⇒ tan(θ +2θ )=√3

True

Explanation:

We have, tan(π cos θ) = cot (π sin θ)

(a) sin (x + y) sin(x – y)

= (sin x.cos y + sin y.cos x)( sin x.cos y - sin y.cos x)

= sin²x cos²y+ sin y.cos x. sin x.cos y- sin x.cos y sin y.cos x- sin²y cos²x

= sin²x cos²y- sin²y cos²x

= sin²x(1- sin²y)- sin²y(1- sin²x)

= sin²x- sin²x sin²y- sin²y+ sin²x sin²y

= sin²x – sin²y

⇒ (a) matches with (iv)

(b) cos(x + y) cos(x – y)

= (cos x.cos y−sin x.sin y)( cos x.cosy+sin x.siny)

= (cos²x cos²y+ cos x.cos y sin x.siny- sin x.sin y cos x.cosy- sin²x sin²y)

= cos²x cos²y- sin²x sin²y

= cos²x(1- sin²y)-(1- cos²x) sin²y

= cos²x- cos²x sin²y- sin²y+ cos²x sin²y

= cos²x – sin²y

⇒ (b) matches with (i)

(c)

⇒ (c) matches with (ii)

(c)

⇒ (d) matches with (iii)