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Given:

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+....}}}}=cosecθ$

Formula used:

For infinite nested radicals:

$\sqrt{k+\sqrt{k+\sqrt{k+....}}} = x \text{ where } x^2 = k+x$

Also, $cosec \theta = \frac{1}{\sin \theta}$

Calculation:

$\sqrt{2+\sqrt{2+\sqrt{2+....}}} = x$

⇒ $cosec \theta = x$

⇒ $cosec \theta = x \text{ and } x^2 = 2+x$

⇒$cosec \theta = x$, and solving $x^2 - x - 2 = 0$

⇒ (x-2)(x+1) = 0 \Rightarrow x = 2 \text{ (taking positive root as \(cosec \theta > 0)}\)

⇒ $cosec \theta = 2$

⇒ $\sin \theta = \frac{1}{cosec \theta} = \frac{1}{2}$

∴ The correct answer is option (4).

Given:

8 sinθ − cosθ = 4, 0 < θ < π/2

Formula used:

cosθ = 8 sinθ − 4; sin²θ + cos²θ = 1; solve quadratic in sinθ

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

Calculations:

⇒ cosθ = 8 sinθ − 4

⇒ Since sin²θ + cos²θ = 1 

⇒ sin²θ + (8 sinθ − 4)² = 1

⇒ sin²θ + 64 sin²θ − 64 sinθ + 16 = 1

⇒ 65 sin²θ − 64 sinθ + 15 = 0

⇒ Discriminant = 64² − 4 × 65 × 15 = 196, √196 = 14

⇒ sinθ = (64 ± 14)/130

⇒ sinθ = 78/130 = 3/5 (valid)

⇒ sinθ = 50/130 = 5/13 (invalid, cosθ negative)

⇒ Valid sinθ = 3/5 ⇒ cosecθ = 1 / sinθ = 5/3

∴ cosecθ = 5/3.

Given:

2 tanθ = sec² θ - 2 , 0 < θ < π/2

Formula used:

sec² θ = 1 + tan² θ

cotθ = 1 / tanθ

Calculations:

⇒ 2 tanθ = sec² θ - 2

⇒ 2 tanθ = (1 + tan² θ) - 2

⇒ 2 tanθ = tan² θ - 1

⇒ tan² θ - 2 tanθ - 1 = 0

Let t = tanθ

⇒ t² - 2t - 1 = 0

⇒ t = (2 ± √(4 + 4)) / 2 = (2 ± √8)/2 = 1 ± √2

In 0 < θ < π/2, tanθ > 0 ⇒ t = 1 + √2

⇒ cotθ = 1 / (1 + √2)

⇒ cotθ = √2 - 1 (after rationalisation)

∴ Correct option: √2 - 1.

Given:

$(\sec \theta - \tan \theta) - \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}}$

Formula used:

$\sec^2 \theta - \tan^2 \theta = 1$

 $\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}} = \sec \theta - \tan \theta$

Calculation:

$(\sec \theta - \tan \theta) - \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}}$

⇒ $(\sec \theta - \tan \theta) - (\sec \theta - \tan \theta)$

⇒ 0

∴ The correct answer is option (1).

Given:

cot θ = √7

Formula used:

cosec²θ = 1 + cot²θ

sec²θ = 1 + tan²θ

tan θ = 1 / cot θ

Calculation:

⇒ tan θ = 1 / √7

⇒ sec²θ = 1 + 1/7 = 8/7

⇒ cosec²θ = 1 + 7 = 8

⇒ Numerator = 8 − 8/7 = (56/7 − 8/7) = 48/7

⇒ Denominator = 8 + 8/7 = (56/7 + 8/7) = 64/7

⇒ Value = (48/7) ÷ (64/7)

⇒ 48/64 = 3/4

∴ The correct answer is 3/4.

Given:

The difference between the two acute angles in a right-angled triangle is $\dfrac{\pi}{12}$ radian.

One of the acute angles of the triangle is to be determined.

Formula used:

In a right-angled triangle:

$\text{Sum of acute angles} = \dfrac{\pi}{2}$

Let one angle be $\theta$ and the other be $\theta - \dfrac{\pi}{12}$.

$\theta + \left(\theta - \dfrac{\pi}{12}\right) = \dfrac{\pi}{2}$

Calculations:

$\theta + \theta - \dfrac{\pi}{12} = \dfrac{\pi}{2}$

⇒ $\theta = \dfrac{\pi}{4} + \dfrac{\pi}{24}$

⇒ $\theta = \dfrac{6\pi + \pi}{24}$

⇒ $\theta = \dfrac{7\pi}{24}$

Converting $\theta$ to degrees:

$\theta = \dfrac{7 \times 180}{24} = 52.5^\circ$

∴ The correct answer is option (3): 52.5º.

Given:

$\log_{10}\big[998+\sqrt{x^2-18x+76}\big]=3$

Formula used:

If log₁₀(A) = 3 ⇒ A = 10³

Calculations:

⇒ 998 + √(x² - 18x + 76) = 10³ = 1000

⇒ √(x² - 18x + 76) = 1000 - 998 = 2

⇒ x² - 18x + 76 = 2² = 4

⇒ x² - 18x + 72 = 0

⇒ x2 - 12x - 6x + 72 = 0

⇒ (x - 12) (x - 6) = 0

⇒ x = 12 or x = 6

⇒ Let α = 12, β = 6

⇒ (α - β)² = (12 - 6)² = 6² = 36

∴ (α - β)² = 36.

Given:

x⁴ + y⁴ = 14x²y²

Formula used:

x⁴ + y⁴ = (x² + y²)² − 2x²y²

log a + log b = log(ab)

Calculation:

x⁴ + y⁴ = (x² + y²)² − 2x²y²

⇒ (x² + y²)² − 2x²y² = 14x²y²

⇒ (x² + y²)² = 16x²y²

⇒ x² + y² = 4xy

Check Statement I:

log(x² + y²) = log(4xy)

⇒ = log4 + logx + logy

⇒ = 2log2 + logx + logy 

Statement I is correct

Check Statement II:

x² − y² = (x − y)(x + y)

From x² + y² = 4xy

⇒ (x − y)² = 2xy

⇒ x − y = √(2xy)

x + y = √(x² + y² + 2xy)

⇒ x + y = √(6xy)

⇒ x² − y² = √(12x²y²)

⇒ = 2√3 xy

log(x² − y²) = log(2√3 xy)

⇒ logx + logy + log2 + 0.5log3 

Statement II is correct

∴ Both I and II are correct.

Given:

(3x + y)² + (3x + y)(x + 5y) − 20(x + 5y)²

Formula used:

a² + ab − 20b² = (a + 5b)(a − 4b)

Calculation:

Let a = (3x + y) and b = (x + 5y)

⇒ Given expression = a² + ab − 20b²

⇒ (a + 5b)(a − 4b)

⇒ [(3x + y) + 5(x + 5y)] × [(3x + y) − 4(x + 5y)]

⇒ (3x + y + 5x + 25y) × (3x + y − 4x − 20y)

⇒ (8x + 26y) × (−x − 19y)

⇒ 2(4x + 13y) × (−1)(x + 19y)

⇒ Factors are (4x + 13y) and (x + 19y)

∴ The correct answer is Both I and II.

Given:

$\dfrac{\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}}{\dfrac{x+y}{x-y}+\dfrac{y+z}{y-z}+\dfrac{z+x}{z-x}+3}$ = ?

Formula used:

y = x - (x - y) ⇒ y/(x - y) = x/(x - y) - 1 

Calculations:

Let A = $\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}$

Then sum $\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x}$ =

⇒ using y/(x-y) = x/(x-y) - 1:

⇒ $\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x} = \dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x} - 3$

So denominator D = $\dfrac{x+y}{x-y}+\dfrac{y+z}{y-z}+\dfrac{z+x}{z-x}+3$ =

⇒ $\Big(\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}\Big) + \Big(\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x}\Big) + 3$

⇒ substitute previous result:

⇒ D = A + (A - 3) + 3 = 2A

Therefore the whole expression =

⇒ $\dfrac{A}{2A} = \dfrac{1}{2}$

∴ The value of the given expression is 1/2.