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Given:

2 sin⁴(α) + 2 cos⁴(α) - 1 = 0, where 0 < α < π/2

Formula used:

sin⁴(α) + cos⁴(α) = (sin²(α) + cos²(α))² - 2sin²(α) cos²(α)

We know that sin²(α) + cos²(α) = 1.

Calculation:

The given equation is:

2 sin⁴(α) + 2 cos⁴(α) - 1 = 0

⇒ 2 (sin⁴(α) + cos⁴(α)) = 1

⇒ sin⁴(α) + cos⁴(α) = 1/2

Now, using the identity:

sin⁴(α) + cos⁴(α) = (sin²(α) + cos²(α))² - 2sin²(α) cos²(α)

⇒ (1)² - 2sin²(α) cos²(α) = 1/2

⇒ 1 - 2sin²(α) cos²(α) = 1/2

⇒ 2sin²(α) cos²(α) = 1/2

⇒ sin²(α) cos²(α) = 1/4

Now, take the square root of both sides:

⇒ sin(α) cos(α) = 1/2

Using the identity sin(2α) = 2 sin(α) cos(α):

⇒ sin(2α) = 2 × 1/2 = 1

sin(2α) = 1 implies that: 2α = 90°

Therefore: α = 45°

Now, find sin(2α) + cos(2α):

Since α = 45°:

sin(2α) = sin(90°) = 1

cos(2α) = cos(90°) = 0

Thus: sin(2α) + cos(2α) = 1 + 0 = 1.

sin(2α) + cos(2α) = 1.

∴ The correct answer is option 2.

Given:

(I): 1 − sin⁶α = cos²α (cos⁴α − 3cos²α + 3)

(II): cos⁸α − sin⁸α = 2sin²α (1 − cos⁴α + sin²α cos²α)

Check (I):

LHS = 1 − sin⁶α = (1 − sin²α)(1 + sin²α + sin⁴α)

⇒ = cos²α (1 + (1 − cos²α) + (1 − cos²α)²)

⇒ = cos²α (3 − 3cos²α + cos⁴α)

⇒ RHS ✔

So, (I) is true.

Check (II):

LHS = cos⁸α − sin⁸α = (cos⁴α − sin⁴α)(cos⁴α + sin⁴α)

⇒ (cos²α − sin²α)(cos²α + sin²α)(cos⁴α + sin⁴α)

⇒ (cos 2α)(1)(cos⁴α + sin⁴α)

RHS = 2sin²α(1 − cos⁴α + sin²α cos²α)

This form does not simplify to LHS.

So, (II) is false.

Final Answer:

Only (I) is an identity.

Given:

p = 1/(cosecθ + cotθ)

q = cosecθ

Calculations:

We have,

p = 1/(cosecθ + cotθ) = 1/ (1/sinθ + cosθ/sinθ)

⇒ 1/{(1 + cosθ)/sinθ}

⇒ sinθ / (1 + cos θ)

q = cosec θ = 1/sinθ

So, p² - 2pq

⇒ {sinθ / (1 + cos θ)}² - 2 × {sinθ / (1 + cos θ)} × {1/sinθ}

⇒ {sin²θ / (1 + cos θ)2} - 2 / (1 + cos θ)

⇒ {sin²θ / (1 + cos θ)2 - 2 / (1 + cos θ)

⇒ (1 - cos²θ) /(1 + cos θ)2 - 2 / (1 + cos θ)

⇒ (1 - cosθ)(1 + cosθ) /(1 + cos θ)2 - 2 / (1 + cos θ)

⇒ (1 - cosθ) / (1 + cos θ) - 2 / (1 + cos θ)

⇒ (1 - cosθ - 2) / (1 + cos θ)

⇒ (-1 - cosθ) / (1 + cos θ)

⇒ - (1 + cosθ) / (1 + cos θ)

⇒ - 1

∴ The correct answer is option (1)

Given:

Statement I: (cosec α - sec α) is always positive in the first quadrant.

Statement II: (tan α - cot α) is always negative in the first quadrant.

Formula used:

1. In the first quadrant, all trigonometric functions are positive.
2. cosec α = 1/sin α, sec α = 1/cos α, tan α = sin α/cos α, cot α = cos α/sin α.

Calculation:

For Statement I:

(cosec α - sec α) = 1/sin α - 1/cos α

⇒ Taking LCM:

(cosec α - sec α) = (cos α - sin α)/(sin α × cos α)

Since cos α and sin α are both positive in the first quadrant, the numerator (cos α - sin α) can be positive, zero, or negative depending on the specific value of α. Thus, Statement I is not always true.

For Statement II:

(tan α - cot α) = (sin α/cos α) - (cos α/sin α)

⇒ Taking LCM:

(tan α - cot α) = (sin² α - cos² α)/(sin α × cos α)

In the first quadrant, sin α and cos α are positive. However, sin² α - cos² α is negative because cos α > sin α for α in the first quadrant. Hence, (tan α - cot α) is always negative in the first quadrant.

∴ The correct answer is option (4).

Given:

AB = h

Calculation:

Let OP be the height of the tower,

In Δ ABP,

tan 30° = p/b = AB / AP

⇒ 1/√3 = h / AP

⇒ AP = h √3

Now, In Δ AOP,

tan 60° = p/b = OP/AP

⇒ √3 = OP / h √3

⇒ OP = h √3 × √3

⇒ OP = 3 h

Hence, the height of the tower is 3h.

Given:

Expression = (sin θ) / (1 − cot θ) + (cos θ) / (1 − tan θ), where θ ≠ π/4.

Formula used:

cot θ = cos θ / sin θ, tan θ = sin θ / cos θ.

Simplify each term separately and then add.

Calculation:

First term:

(sin θ) / (1 − cot θ) = (sin θ) / (1 − cos θ/sin θ)

= (sin θ) / ((sin θ − cos θ)/sin θ)

= (sin²θ) / (sin θ − cos θ)

Second term:

(cos θ) / (1 − tan θ) = (cos θ) / (1 − sin θ/cos θ)

= (cos θ) / ((cos θ − sin θ)/cos θ)

= (cos²θ) / (cos θ − sin θ)

= − (cos²θ) / (sin θ − cos θ)

Now, add both terms:

(sin²θ)/(sin θ − cos θ) − (cos²θ)/(sin θ − cos θ)

= (sin²θ − cos²θ)/(sin θ − cos θ)

= −(cos²θ − sin²θ)/(sin θ − cos θ)

= −(cos θ − sin θ)(cos θ + sin θ)/(sin θ − cos θ)

= −(−1)(cos θ + sin θ)

= cos θ + sin θ

∴ The correct answer is (sin θ + cos θ).

Given:

Radius (r) = 4 cm

Length of arc (l) = π cm

Formula used:

Length of arc (l) = r × θ

Where θ = Angle subtended by the arc at the centre (in radians)

Calculation:

l = r × θ

π = 4 × θ

⇒ θ = π / 4

Hence, the magnitude of the angle subtended by the arc at the center is π/4.

The correct answer is option (4).

Given:

cot2θ - $3\sqrt{3}$ cotθ + 6 = 0

Formula used:

cot2θ - $3\sqrt{3}$ cotθ + 6 = 0

put cotθ = x

So, x² - $3\sqrt{3}$x + 6 = 0

⇒ x² - 2$\sqrt3$x - $\sqrt3$x + 6 = 0

⇒ x(x - 2$\sqrt3$) - $\sqrt3$(x - 2$\sqrt3$) = 0

⇒ (x - 2$\sqrt3$) (x - $\sqrt3$) = 0

so, x = $\sqrt3$ & 2$\sqrt3$

i.e. cot θ = $\sqrt3$ = cot 30°

So, θ = 30°

Now,

sin θ + cos 2θ

⇒ sin 30 + cos 60

⇒ 1/2 + 1/2 = 1

∴ The correct answer is option (2).

Given:

I. sin²θ = ((x + y)²) / (4xy), where x, y are positive unequal real numbers.

II. sin θ + cos θ = x + 1/x, where x is a positive real number.

Formula used:

For any real angle θ, sin²θ ≤ 1 and |sin θ + cos θ| ≤ √2.

Check whether the given equations satisfy these constraints.

Calculation:

For I: sin²θ = ((x + y)²) / (4xy)

Since x, y > 0 and x ≠ y, (x + y)² > 4xy ⇒ ((x + y)²)/(4xy) > 1

But sin²θ ≤ 1 always.

Hence equation I is not possible.

For II: sin θ + cos θ = x + 1/x

By AM–GM inequality, x + 1/x ≥ 2 for x > 0

Also, |sin θ + cos θ| ≤ √2 ≈ 1.414

So x + 1/x ≥ 2 > √2 ⇒ equation II is also not possible.

∴ Neither I nor II is possible.

Given:

Equation: m²(sin θ - 1) + n²(sin θ + 1) = 0

Range: 0 ≤ θ ≤ π/2

Expression to find: (m² + n²) cos θ - (m² - n²) cot θ

Formula used:

Basic trigonometric identities:

cot θ = cos θ / sin θ

Calculation:

From the given equation:

⇒ m²(sin θ - 1) + n²(sin θ + 1) = 0

⇒ m² sin θ - m² + n² sin θ + n² = 0

⇒ (m² + n²) sin θ + (-m² + n²) = 0

⇒ (m² + n²) sin θ = (m² - n²)

⇒ sin θ = (m² - n²) / (m² + n²)

Now, substitute sin θ into the given expression:

(m² + n²) cos θ - (m² - n²) cot θ

⇒ (m² + n²) cos θ - (m² - n²) × (cos θ / sin θ)

⇒ (m² + n²) cos θ - (m² - n²) × (cos θ / [(m² - n²) / (m² + n²)])

⇒ (m² + n²) cos θ - (m² - n²) × (cos θ × (m² + n²) / (m² - n²))

⇒ (m² + n²) cos θ - (m² + n²) cos θ

⇒ 0

∴ The correct answer is option (4).