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sin α + cos α = $\sqrt{2}$

It is valid when α = π/4

As, sin π/4 + cos π/4 = 1/$\sqrt{2}$ + 1/$\sqrt{2}$ = 2/$\sqrt{2}$ = $\sqrt{2}$

So, sin α - cos α = sin π/4 - cos π/4 = 1/$\sqrt{2}$ - 1/$\sqrt{2}$ = 0

So, sin3α − cos3α

⇒ (sin α - cos α) (sin² α + cos²α - sinα cosα)

⇒ 0 (sin2 α + cos2α - sinα cosα)

⇒ 0

Hence, the correct answer is 0.

Alternate Method

Given:

sin α + cos α = √2, where 0 < α < π/2

Formula used:

sin³α - cos³α = (sin α - cos α)(sin²α + sin α cos α + cos²α)

Calculation:

sin α + cos α = √2

⇒ (sin α + cos α)² = (√2)²

⇒ sin²α + cos²α + 2sin α cos α = 2

⇒ 1 + 2sin α cos α = 2

⇒ 2sin α cos α = 1

⇒ sin α cos α = 1/2

Now, sin α - cos α = √(sin²α + cos²α - 2sin α cos α)

⇒ sin α - cos α = √(1 - 1)

⇒ sin α - cos α = 0

Plugging into the formula:

sin³α - cos³α = (sin α - cos α)(sin²α + sin α cos α + cos²α)

⇒ sin³α - cos³α = 0 × (1 + 1/2)

⇒ sin³α - cos³α = 0

∴ The correct answer is option (4).

Given:

Expression: (1 + cot α − cosec α)(1 + tan α + sec α)

Formula used:

cot α = cos α / sin α

cosec α = 1 / sin α

tan α = sin α / cos α

sec α = 1 / cos α

(x − 1)(x + 1) = x² − 1

sin²α + cos²α = 1

Calculations:

(1 + cot α − cosec α)(1 + tan α + sec α)

⇒ (1 + cos α / sin α − 1 / sin α)(1 + sin α / cos α + 1 / cos α)

⇒ ((sin α + cos α − 1) / sin α) × ((cos α + sin α + 1) / cos α)

Multiply numerators:

⇒ ((sin α + cos α − 1)(sin α + cos α + 1)) / (sin α × cos α)

Use difference of squares:

⇒ ((sin α + cos α)² − 1) / (sin α × cos α)

Expand and apply identity:

⇒ (sin²α + cos²α + 2 sin α cos α − 1) / (sin α × cos α)

⇒ (1 + 2 sin α cos α − 1) / (sin α × cos α)

⇒ 2 sin α cos α / (sin α × cos α)

⇒ 2

∴ (1 + cot α − cosec α)(1 + tan α + sec α) = 2

Given:

tan θ = (sin α - cos α) / (sin α + cos α)

θ and α are acute angles, α ≠ π/4.

Formula used:

√(2)sin θ = ?

Calculation:

tan θ = (sin α - cos α) / (sin α + cos α)

⇒ θ = tan^-1[(sin α - cos α) / (sin α + cos α)]

Using trigonometric identity: sin θ = tan θ / √(1 + tan²θ)

⇒ sin θ = [(sin α - cos α) / (sin α + cos α)] / √(1 + [(sin α - cos α) / (sin α + cos α)]²)

Let k = sin α - cos α and m = sin α + cos α.

⇒ sin θ = k / m / √[1 + (k / m)²]

⇒ sin θ = k / √(m² + k²)

⇒ √(2)sin θ = √(2) × (sin α - cos α) / √[(sin α + cos α)² + (sin α - cos α)²]

⇒ √(2)sin θ = √(2) × (sin α - cos α) / √[2(sin²α + cos²α)]

⇒ √(2)sin θ = √(2) × (sin α - cos α) / √(2)

⇒ √(2)sin θ = sin α - cos α

∴ The correct answer is option (1).

Given:

(sin α + 2)(sin α + 4)(sin α − 2)(sin α − 4) = 0

Formula used:

Expression = 0 ⇔ at least one factor = 0. But sin α ∈ [−1, 1]

Calculations:

⇒ sin α + 2 = 0 ⇒ sin α = −2 (not possible)

⇒ sin α + 4 = 0 ⇒ sin α = −4 (not possible)

⇒ sin α − 2 = 0 ⇒ sin α = 2 (not possible)

⇒ sin α − 4 = 0 ⇒ sin α = 4 (not possible)

Number of real values of α = 0.

∴ The correct answer is option 1.

Given:

tan3x × tan6x = 1, 0 ≤ x ≤ 30°

Formula used:

tanA × tanB = 1 ⇒ tanA = cotB = tan(90° − B + k × 180°)

Calculations:

⇒ tan3x = tan(90° − 6x + n × 180°)

⇒ 3x = 90° − 6x + n × 180°

⇒ 9x = 90° + n × 180°

⇒ x = 10° + n × 20°

Check n values for 0 ≤ x ≤ 30°:

n = 0 → x = 10° (valid)

n = 1 → x = 30° (but 3x = 90° → tan3x undefined → reject)

n = −1 → x = −10° (out of range)

∴ x = 10°.

Given:

Expression to evaluate:

$\frac{(a - b)^2}{(b - c)(c - a)} + \frac{(b - c)^2}{(c - a)(a - b)} + \frac{(c - a)^2}{(a - b)(b - c)} - 3$

Step 1: Let

x = $\frac{(a - b)^2}{(b - c)(c - a)}$

y = $\frac{(b - c)^2}{(c - a)(a - b)}$

z = $\frac{(c - a)^2}{(a - b)(b - c)}$

So expression becomes: x + y + z − 3

Step 2: Take LCM of all 3 terms:

LCM = (a − b)(b − c)(c − a)

Write all terms with common denominator:

x = $\frac{(a - b)^2}{(b - c)(c - a)} = \frac{(a - b)^3}{(a - b)(b - c)(c - a)}$

y = $\frac{(b - c)^2}{(c - a)(a - b)} = \frac{(b - c)^3}{(b - c)(c - a)(a - b)}$

z = $\frac{(c - a)^2}{(a - b)(b - c)} = \frac{(c - a)^3}{(c - a)(a - b)(b - c)}$

Numerators:

⇒ x + y + z = $\frac{(a - b)^3 + (b - c)^3 + (c - a)^3}{(a - b)(b - c)(c - a)}$

Use identity:

If x + y + z = 0 ⇒ x³ + y³ + z³ = 3xyz

Here, let: u = (a − b), v = (b − c), w = (c − a)

Then: u + v + w = 0

So: u³ + v³ + w³ = 3uvw

Apply:

Numerator = (a − b)³ + (b − c)³ + (c − a)³ = 3(a − b)(b − c)(c − a)

Denominator = (a − b)(b − c)(c − a)

So: x + y + z = $\frac{3(a - b)(b - c)(c - a)}{(a - b)(b - c)(c - a)} = 3$

Final Expression = x + y + z − 3 = 3 − 3 = 0

∴ The correct answer is: 0

Given:

Expression: $\dfrac{100!}{100^n}$ is an integer

Formula used:

We need to find the largest n such that 100ⁿ divides 100!

100 = 2² × 5² ⇒ 100ⁿ = 2²ⁿ × 5²ⁿ

So, we must find maximum n such that both 2²ⁿ and 5²ⁿ divide 100!

Calculations:

Powers of 2 in 100!

⇒ [100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64]

⇒ 50 + 25 + 12 + 6 + 3 + 1 = 97

Powers of 5 in 100!

⇒ [100/5] + [100/25] + [100/125]

⇒ 20 + 4 + 0 = 24

Now: 2²ⁿ must ≤ 2⁹⁷ ⇒ 2n ≤ 97 ⇒ n ≤ 48.5

And: 5²ⁿ must ≤ 5²⁴ ⇒ 2n ≤ 24 ⇒ n ≤ 12

So, the limiting factor is powers of 5 ⇒ n ≤ 12

The largest value of n is 12.

∴ The correct answer is option 4.

Given:

Distance traveled East = x meters

Distance traveled South = 40 meters

Distance traveled West = 60 meters

Distance traveled North = 200 meters

PR = 200 meters

Formula used:

Pythagoras theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Calculation:

PR = 200 meters

Horizontal distance between P and R = x - 60

Vertical distance between P and R = 200 - 40 = 160

Using Pythagoras theorem:

⇒ (PR)² = (Horizontal distance)² + (Vertical distance)²

⇒ 200² = (x - 60)² + 160²

⇒ 40000 = (x - 60)² + 25600

⇒ (x - 60)² = 40000 - 25600

⇒ (x - 60)² = 14400

⇒ x - 60 = √14400

⇒ x - 60 = 120

⇒ x = 120 + 60

⇒ x = 180

∴ The correct answer is option (2).

Given:

Equation: x² + y² + z² = 3

x, y, and z are integers.

Formula used:

The algebraic identity: (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

Calculations:

We are given x² + y² + z² = 3, where x, y, and z are integers.

Since x, y, and z are integers, their squares (x², y², z²) must be non-negative integers (0, 1, 4, 9, ...).

The only way to sum three non-negative integer squares to get 3 is if each square is 1

⇒ x² = 1, y² = 1, z² = 1

This implies that x, y, and z can only be either 1 or -1.

Now, let P = (xy + yz + zx) and S = (x + y + z).

Using the algebraic identity:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

⇒ S² = 3 + 2P

Rearrange the equation to solve for P:

⇒ 2P = S² − 3

⇒ P = $(\dfrac{S^2 - 3}{2})$

Now, we will consider all possible combinations of x, y, and z from {1, -1} and calculate S and P for each case:

Case 1: All three integers are 1.

(x, y, z) = (1, 1, 1)

S = 1 + 1 + 1 = 3

P = $(\dfrac{3^2 - 3}{2})$ = $(\dfrac{9 - 3}{2})$ = $(\dfrac{6}{2})$ = 3

Case 2: Two integers are 1, and one is -1.

Possible combinations: (1, 1, -1), (1, -1, 1), (-1, 1, 1)

For any of these combinations, S = 1 + 1 − 1 = 1

P = $(\dfrac{1^2 - 3}{2})$ = $(\dfrac{1 - 3}{2})$ = $(\dfrac{-2}{2})$ = -1

Case 3: One integer is 1, and two are -1.

Possible combinations: (1, -1, -1), (-1, 1, -1), (-1, -1, 1)

For any of these combinations, S = 1 − 1 − 1 = -1

P = $(\dfrac{(-1)^2 - 3}{2})$ = $(\dfrac{1 - 3}{2})$ = $(\dfrac{-2}{2})$ = -1

Case 4: All three integers are -1.

(x, y, z) = (-1, -1, -1)

S = -1 − 1 − 1 = -3

P = $(\dfrac{(-3)^2 - 3}{2})$ = $(\dfrac{9 - 3}{2})$ = $(\dfrac{6}{2})$ = 3

The distinct values that (xy + yz + zx) can have are 3 and -1.

Therefore, there are 2 possible values for (xy + yz + zx).

Given:

x + y + z = 10

xy + yz + zx = 18

Formula used:

The expression for the sum of cubes of three numbers can be expressed as:

x³ + y³ + z³ - 3xyz = (x + y + z) [(x + y + z)² - 3(xy + yz + zx)]

Calculation:

Substituting the given values:

x³ + y³ + z³ - 3xyz = (10) [(10)² - 3(18)]

⇒ x³ + y³ + z³ - 3xyz = 10 [100 - 54]

⇒ x³ + y³ + z³ - 3xyz = 10 × 46

⇒ x³ + y³ + z³ - 3xyz = 460

∴ The correct answer is option (3).