If tan φ + cot φ = 5, then find the value of (tan2 φ + cot2 φ).
The problem provides an equation involving \(\tan \phi\) and \(\cot \phi\) and asks for the value of an expression involving \(\tan^2 \phi\) and \(\cot^2 \phi\). We can use algebraic identities to solve this. Specifically, the identity \((a+b)^2 = a^2 + b^2 + 2ab\) will be useful.
Given the equation:
\[ \tan \phi + \cot \phi = 5 \]We need to find the value of \(\tan^2 \phi + \cot^2 \phi\). Let's square both sides of the given equation:
\[ (\tan \phi + \cot \phi)^2 = 5^2 \]Expand the left side using the algebraic identity \((a+b)^2 = a^2 + b^2 + 2ab\), where \(a = \tan \phi\) and \(b = \cot \phi\):
\[ \tan^2 \phi + \cot^2 \phi + 2(\tan \phi)(\cot \phi) = 25 \]Recall that \(\cot \phi = \frac{1}{\tan \phi}\). Therefore, \[(\tan \phi)(\cot \phi) = (\tan \phi)\left(\frac{1}{\tan \phi}\right) = 1\].
Substitute this value back into the equation from Step 3:
\[ \tan^2 \phi + \cot^2 \phi + 2(1) = 25 \] \[ \tan^2 \phi + \cot^2 \phi + 2 = 25 \]Subtract 2 from both sides to find the required value:
\[ \tan^2 \phi + \cot^2 \phi = 25 - 2 \] \[ \tan^2 \phi + \cot^2 \phi = 23 \]Correct Option: D) 23
The question asks to identify the substance where carbon atoms are arranged in a football-like shape. This refers to a specific allotrope of carbon known for its unique cage-like structure.
A) Fullerenes — Fullerenes are allotropes of carbon characterized by their cage-like structures. The most well-known fullerene, C60 (Buckminsterfullerene), has 60 carbon atoms arranged in a spherical shape resembling a football, with 12 pentagonal and 20 hexagonal faces. This perfectly matches the description in the question.
The question asks to identify the author of the famous Sanskrit epic 'Raghuvamsham'. This requires knowledge of prominent ancient Indian poets and their works.
Correct Option: C) Kalidas
The question asks for the melting point of ice. This is a fundamental concept in physics and chemistry, related to temperature scales. We need to recall the standard melting point of ice in different temperature scales, specifically Celsius and Kelvin, and then identify the correct option.
Correct Option: B) 273 K
The melting point of ice is \(0^\circ \text{C}\), which is equivalent to \(273.15 \text{ K}\) (often rounded to \(273 \text{ K}\)).
The question asks for the maximum permissible strength of the Rajya Sabha, which is the upper house of the Indian Parliament. This is a factual question related to the Indian Constitution and parliamentary structure.
Correct Option: A) 250 is the maximum permissible strength of the Rajya Sabha as stipulated by the Indian Constitution.
The problem presents a statement about beggars on footpaths at tourist places and two courses of action. We need to evaluate each course of action to determine if it logically follows as a reasonable and effective response to the problem described.
Correct Option: A) Both 1 and 2 follow
Both courses of action are logically sound. Course of action 1 provides an immediate, albeit temporary, solution to clear the footpaths and manage the situation. Course of action 2 provides a long-term, humanitarian solution by addressing the root causes of begging through rehabilitation and social upliftment. A comprehensive approach often requires both immediate intervention and long-term strategic planning.
The question asks to identify the water body that lies between Australia and New Zealand. This requires geographical knowledge of the Oceania region.
B) Tasman Sea — The Tasman Sea is a marginal sea of the South Pacific Ocean, situated between Australia and New Zealand. It is approximately 2,000 kilometers (1,200 mi) wide and is named after the Dutch explorer Abel Tasman, who was the first known European to reach New Zealand and Tasmania.
From the top of a 150 m tall building A, the angle of elevation to the top of the building B is 45 degrees and the angle of depression to the bottom of the building B is 30 degrees. What is the height of the building B?
250 m
\(\frac{450}{\sqrt{3}}\) m
\(\frac{150}{\sqrt{3}}\) m
\(\displaystyle 150(1+\sqrt{3})\) m
This problem involves trigonometry, specifically angles of elevation and depression. We need to use the tangent function to relate the angles to the heights and the horizontal distance between the buildings. We will set up two right-angled triangles and solve for the unknown height of building B.
Let the height of building A be \(h_A = 150 \text{ m}\). Let the height of building B be \(h_B\). Let the horizontal distance between the buildings be \(x\).
From the top of building A, the angle of depression to the bottom of building B is \(30^\circ\). Let's consider the right-angled triangle formed by the top of building A, the bottom of building B, and the point on the ground directly below the top of building A. In this triangle, the opposite side is \(h_A\) and the adjacent side is \(x\).
\[ \tan(30^\circ) = \frac{h_A}{x} \]\[ \frac{1}{\sqrt{3}} = \frac{150}{x} \]\[ x = 150\sqrt{3} \text{ m} \]From the top of building A, the angle of elevation to the top of building B is \(45^\circ\). Let \(h'\) be the height difference between the top of building B and the top of building A. So, \(h' = h_B - h_A\). Consider the right-angled triangle formed by the top of building A, the top of building B, and the point on building B at the same height as the top of building A. In this triangle, the opposite side is \(h'\) and the adjacent side is \(x\).
\[ \tan(45^\circ) = \frac{h'}{x} \]\[ 1 = \frac{h'}{x} \]\[ h' = x \]Substitute the value of \(x\) from step 2 into the equation from step 3:
\[ h' = 150\sqrt{3} \text{ m} \]Now, find the total height of building B:
\[ h_B = h_A + h' \]\[ h_B = 150 + 150\sqrt{3} \]\[ h_B = 150(1 + \sqrt{3}) \text{ m} \]D) \(150(1+\sqrt{3})\) m. This matches our calculated height for building B.
The problem asks for the angle between force and displacement when the work done is zero. We need to recall the formula for work done in physics, which involves the dot product of force and displacement vectors, and then determine the angle that makes the work zero.
D) 90 degrees is correct because when the angle between the force and displacement is \(90^\circ\), \(\cos(90^\circ) = 0\), which makes the work done \(W = Fd \cdot 0 = 0\).