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Explanation:

We are given:

$p = \tan(2\alpha) - \tan(\alpha)$

$q = \cot(\alpha) - \cot(2\alpha)$

We rewrite q  in terms of tangent since $\cot(\theta) = \frac{1}{\tan(\theta)}$

$q = \frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)}$

Now we compute $\cot(\theta) = \frac{1}{\tan(\theta)}$

$\frac{p}{q} = \frac{\tan(2\alpha) - \tan(\alpha)}{\frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)}}$

Next, we find a common denominator for q :

$\frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)} = \frac{\tan(2\alpha) - \tan(\alpha)}{\tan(\alpha) \cdot \tan(2\alpha)}$

Substituting this back into the formula for $\frac{p}{q}$

$\frac{p}{q} = \frac{\tan(2\alpha) - \tan(\alpha)}{\frac{\tan(2\alpha) - \tan(\alpha)}{\tan(\alpha) \cdot \tan(2\alpha)}}$

Simplifying, we get:

$\frac{p}{q} = \tan(\alpha) \cdot \tan(2\alpha)$

∴ The correct answer is Option (c): $\tan(\alpha) \cdot \tan(2\alpha)$

Calculation: 

We are given:

$p = \tan(2\alpha) - \tan(\alpha)$

$q = \cot(\alpha) - \cot(2\alpha)$

We need to find p + q.

$p + q = \left( \frac{\sin^2(2\alpha) - \cos^2(2\alpha)}{\sin(2\alpha) \cdot \cos(2\alpha)} \right) + \left( \frac{\cos^2(\alpha) - \sin^2(\alpha)}{\sin(\alpha) \cdot \cos(\alpha)} \right)$

Simplifying both terms:

$= \frac{-2\cos(4\alpha)}{\sin(4\alpha)} + \frac{2\cos(2\alpha)}{\sin(2\alpha)}$

Now, factorizing and simplifying further:

$= \frac{2(\sin(4\alpha) \cdot \cos(2\alpha) - \cos(4\alpha) \cdot \sin(2\alpha))}{\sin(4\alpha) \cdot \sin(2\alpha)}$

Recognizing the sine identity, we get:

$= \frac{2 \sin(4\alpha - 2\alpha)}{\sin(4\alpha) \cdot \sin(2\alpha)}$

Finally, simplifying this:

$= \frac{2 \sin(2\alpha)}{\sin(4\alpha) \cdot \sin(2\alpha)}$

The final result is:

$p + q = 2 cosec (4\alpha)$

∴ The correct answer is Option (4)

Calculation: 

We are given:

$p = \tan(2\alpha) - \tan(\alpha)$

$q = \cot(\alpha) - \cot(2\alpha)$

We need to find $\tan^2(\alpha)$ in terms of p  and q .

We start by simplifying the expression $\frac{p}{p + 2q}$

$\frac{p}{p + 2q} = \frac{1}{1 + \frac{2q}{p}}$

This simplifies further as:

$\frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}}$

Now, simplify the fraction inside the denominator:

$= \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}} = \frac{1}{1 + \frac{1}{2} \cdot \tan(\alpha) \cdot \tan(2\alpha)}$

Now, expand both terms using trigonometric identities:

$= \frac{\sin(\alpha) \cdot \sin(2\alpha)}{\sin(\alpha) \cdot \cos(\alpha) + \cos(\alpha) \cdot \cos(2\alpha)}$

Simplifying this expression gives:

$= \tan^2(\alpha)$

∴ The correct answer is Option (c):

Calculation: 

We are given:

$2\sin(\alpha) + \cos(\alpha) = 2$

$\cos(\alpha) = 2(1 - \sin(\alpha))$

$\cos^2(\alpha) = 4(1 - \sin(\alpha))^2$

Use the identity $\cos^2(\alpha) = 1 - \sin^2(\alpha)$

$1 - \sin^2(\alpha) = 4(1 - 2\sin(\alpha) + \sin^2(\alpha))$

$1 - \sin^2(\alpha) = 4 - 8\sin(\alpha) + 4\sin^2(\alpha)$

Rearrange the terms to form a quadratic equation:

$5\sin^2(\alpha) - 8\sin(\alpha) + 3 = 0$

Using the quadratic formula:

$\sin(\alpha) = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(3)}}{2(5)}$

$\sin(\alpha) = \frac{8 \pm \sqrt{64 - 60}}{10}$

$\sin(\alpha) = \frac{8 \pm 2}{10}$

$\sin(\alpha) = 1$ or $\sin(\alpha) = \frac{3}{5}$

Since $0^\circ < \alpha < 90^\circ$, we select $\sin(\alpha) = \frac{3}{5}$

$\cos^2(\alpha) = 1 - \sin^2(\alpha) = 1 - \left( \frac{3}{5} \right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$

$\cos(\alpha) = \frac{4}{5}$

$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

∴ The correct answer is Option (c):

Calculation: 

We know:

$\sin(\alpha) = \frac{3}{5} \quad \text{and} \quad \cos(\alpha) = \frac{4}{5}$

$\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)$

$\cos(2\alpha) = 1 - 2\sin^2(\alpha)$

$\sin(2\alpha) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$

$\cos(2\alpha) = 1 - 2 \times \left( \frac{3}{5} \right)^2 = 1 - 2 \times \frac{9}{25} = \frac{7}{25}$

$2\sin(2\alpha) + \cos(2\alpha) = 2 \times \frac{24}{25} + \frac{7}{25}$

Simplifying:

$2\sin(2\alpha) + \cos(2\alpha) = \frac{48}{25} + \frac{7}{25} = \frac{55}{25} = \frac{11}{5}$

∴ The correct answer is Option (2):

Calculation:

We are given the equation for the ratio of sides using the Sine Rule:

$\frac{a}{\sin(3x)} = \frac{b}{\sin(x)}$

$\frac{a}{b} = \frac{\sin(3x)}{\sin(x)}$

Step 3: Use the identity for sin(3x), which is $\sin(3x) = 3\sin(x) - 4\sin^3(x)$, and substitute it into the equation:

$2 = \frac{3\sin(x) - 4\sin^3(x)}{\sin(x)}$

$2\sin(x) = 3\sin(x) - 4\sin^3(x)$

$-\sin(x) + 4\sin^3(x) = 0$

$\sin(x)(4\sin^2(x) - 1) = 0$

We have two possible solutions for this equation:

$\sin(x) = 0$, which gives x = $0^\circ$ (not valid in this case).

$4\sin^2(x) - 1 = 0$, which simplifies to:

$\sin^2(x) = \frac{1}{4} \quad \Rightarrow \quad \sin(x) = \frac{1}{2}$

$x = 30^\circ$

∴ The correct answer is Option (2)

Explanation:

We are given the triangle with angles:

$\angle B = x = 30^\circ$

$\angle A = 3x = 90^\circ$

$\angle C = 180^\circ - 120^\circ = 60^\circ$

Step 1: Check if the sum of the angles is 180°:

$\angle A + \angle B + \angle C = 90^\circ + 30^\circ + 60^\circ = 180^\circ$

This confirms that the angles satisfy the angle sum property of a triangle.

Statement I. The triangle is right-angled.

Since$\angle A = 90^\circ$, the triangle is right-angled.

Statement III: III. The angles A, C and B of the triangle are in AP.

The angles $30^\circ 60^\circ , \text and 90^\circ$ are in Arithmetic Progression because:

$60^\circ - 30^\circ = 30^\circ \quad \text{and} \quad 90^\circ - 60^\circ = 30^\circ$

This confirms that the angles are in AP.

Statement II is not correct because there is no mention of a side being 3 times the other.

∴ The correct answer is Option (I) and (III) are correct.

Hence, the correct answer is Option 3.

Given:

• The height of the chimney (PR) is given as 50 meters.
• The man is 100 meters away from the base of the chimney (point P).
• The angle of elevation from the man (at point M) to the highest point of the smoke (point Q) is 45^∘.
• There is a straight line from point P through point R to point Q, perpendicular to PM.

Calculation:

For the right-angled triangle △ MPQ the tangent of the angle of elevation at point M is

⇒ tan 45^∘ = PQ/PM = $\frac{PQ}{100}$ 

Since tan 45^∘ = 1, we have PQ = 100 m.

Finding the angle ∠RMQ:

∠RMQ = ∠PMQ - ∠PMR

In △ PMR, tan ∠PMR = PR/PM = 50/100 = 1/2.

Using the formula tan(A - B) = (tan A - tan B) / (1 + tan A ⋅ tan B):

⇒ tan ∠RMQ = (tan 45^∘ - tan ∠PMR) / (1 + tan 45^∘ ⋅ tan ∠PMR)

⇒ tan ∠RMQ = (1 - 1/2) / (1 + 1 ⋅ 1/2) = (1/2) / (3/2) = 1/3

⇒ ∠RMQ = tan^-1$\frac{1}{3}$

Hence, the Correct answer is Option 2.

Calculation:

We are given that k is a root of x2−4x+1=0

Solving the quadratic equation:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2}$

⇒ $x = 2 \pm \sqrt{3}$

Thus, the two possible values of k are

⇒ $k = 2 + \sqrt{3} \quad \text{or} \quad k = 2 - \sqrt{3}$

We need to find  $\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right)$

Using the identity for the sum of inverse tangents

$\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)$

⇒ $\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - k \cdot \frac{1}{k}}\right)$

$= \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - 1}\right) = \tan^{-1} \left(\frac{k + \frac{1}{k}}{0}\right)$

This expression results in an undefined value, but we know from properties of the inverse tangent that

⇒ $\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \frac{\pi}{2}$

Hence, the correct answer is Option 4.

Calculation:

Given,

$\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4}$

Use the addition formula for inverse tangents:

$\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)$

Using this for $\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right)$, we get:

$\tan^{-1}\left(\frac{k + \frac{1}{2}}{1 - k \cdot \frac{1}{2}}\right) = \frac{\pi}{4}$

Since $\tan\left(\frac{\pi}{4}\right) = 1$, we have:

$\frac{k + \frac{1}{2}}{1 - \frac{k}{2}} = 1$

$k + \frac{1}{2} = 1 - \frac{k}{2}$

$2k + 1 = 2 - k$

$2k + k = 2 - 1$

$3k = 1$

$k = \frac{1}{3}$

Hence, the correct answer is Option 3.