Simplify \(\left(\sec\theta - \frac{1}{\sec\theta}\right) \times \left(\text{cosec}\theta - \frac{1}{\text{cosec}\theta}\right) \times \left(\sin\theta - \frac{1}{\sin\theta}\right).\)
\( \cos ^{3}\theta \)
\( -\sin ^{3}\theta \)
\( \sin ^{3}\theta \)
\( -\cos ^{3}\theta \)
To simplify the given trigonometric expression, we will first convert all terms into their basic sine and cosine forms. Then, we will simplify each bracket individually by finding a common denominator and combining the terms. Finally, we will multiply the simplified expressions from each bracket and use trigonometric identities to reach the final simplified form.
Recall the reciprocal identities:
Substitute these into the given expression:
\[ \left(\frac{1}{\cos\theta} - \cos\theta\right) \times \left(\frac{1}{\sin\theta} - \sin\theta\right) \times \left(\sin\theta - \frac{1}{\sin\theta}\right) \]First bracket:
\[ \frac{1}{\cos\theta} - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} \]Using the identity \( \sin^2\theta + \cos^2\theta = 1 \Rightarrow 1 - \cos^2\theta = \sin^2\theta \), we get:
\[ \frac{\sin^2\theta}{\cos\theta} \]Second bracket:
\[ \frac{1}{\sin\theta} - \sin\theta = \frac{1 - \sin^2\theta}{\sin\theta} \]Using the identity \( \sin^2\theta + \cos^2\theta = 1 \Rightarrow 1 - \sin^2\theta = \cos^2\theta \), we get:
\[ \frac{\cos^2\theta}{\sin\theta} \]Third bracket:
\[ \sin\theta - \frac{1}{\sin\theta} = \frac{\sin^2\theta - 1}{\sin\theta} \]Using the identity \( \sin^2\theta + \cos^2\theta = 1 \Rightarrow \sin^2\theta - 1 = -\cos^2\theta \), we get:
\[ \frac{-\cos^2\theta}{\sin\theta} \]Now, multiply the simplified forms of the three brackets:
\[ \left(\frac{\sin^2\theta}{\cos\theta}\right) \times \left(\frac{\cos^2\theta}{\sin\theta}\right) \times \left(\frac{-\cos^2\theta}{\sin\theta}\right) \]Combine the numerators and denominators:
\[ \frac{\sin^2\theta \times \cos^2\theta \times (-\cos^2\theta)}{\cos\theta \times \sin\theta \times \sin\theta} \] \[ \frac{-\sin^2\theta \cos^4\theta}{\sin^2\theta \cos\theta} \]Cancel \( \sin^2\theta \) from the numerator and denominator:
\[ \frac{-\cos^4\theta}{\cos\theta} \]Cancel \( \cos\theta \) from the numerator and denominator:
\[ -\cos^{4-1}\theta = -\cos^3\theta \]Correct Option: D) \( -\cos^3\theta \)
This problem involves calculating the shortest distance and direction to return to the starting point after a series of movements. We can solve this by breaking down the movements into their North-South and East-West components and then finding the net displacement.
D) 1 km to the east. The final position of Shaan (Point P) is 1 km to the West of Point A. Therefore, to return to Point A, he must drive 1 km to the East.
The question asks about a vitamin deficiency caused by a blocked bile duct. This condition affects the absorption of certain types of vitamins. We need to identify which category of vitamins relies on bile for absorption.
A) Vitamin A — Vitamin A is a fat-soluble vitamin. Its absorption in the small intestine requires the presence of bile. A blocked bile duct prevents bile from reaching the intestine, leading to malabsorption of fats and fat-soluble vitamins, including Vitamin A.
This problem involves compound interest. We are given the total amount of money, the interest rate, and the time periods after which two different investments yield the same final amount. We need to find the initial amount given to one of the sons. We will use the compound interest formula to set up an equation based on the condition that the final amounts are equal.
Let the total amount Shan has be \(P_{\text{total}} = \text{₹}1,617\).
Let the amount given to Piyush be \(P_1\) and the amount given to Manoj be \(P_2\).
We know that \(P_1 + P_2 = \text{₹}1,617\).
The rate of interest is \(R = 10\%\) per annum, compounded annually.
The time for Piyush's investment is \(T_1 = 11\) years.
The time for Manoj's investment is \(T_2 = 12\) years.
The formula for the amount \(A\) after compound interest is given by:
\[ A = P \left(1 + \frac{R}{100}\right)^T \]According to the problem, Piyush and Manoj got the same amount after their respective investment periods. Let this amount be \(A_{\text{final}}\).
For Piyush:
\[ A_{\text{final}} = P_1 \left(1 + \frac{10}{100}\right)^{11} = P_1 \left(1 + \frac{1}{10}\right)^{11} = P_1 \left(\frac{11}{10}\right)^{11} \quad (1) \]For Manoj:
\[ A_{\text{final}} = P_2 \left(1 + \frac{10}{100}\right)^{12} = P_2 \left(1 + \frac{1}{10}\right)^{12} = P_2 \left(\frac{11}{10}\right)^{12} \quad (2) \]Equating (1) and (2):
\[ P_1 \left(\frac{11}{10}\right)^{11} = P_2 \left(\frac{11}{10}\right)^{12} \]Divide both sides by \( \left(\frac{11}{10}\right)^{11} \):
\[ P_1 = P_2 \left(\frac{11}{10}\right) \]\[ P_1 = \frac{11}{10} P_2 \]Now we have a relationship between \(P_1\) and \(P_2\). We also know that \(P_1 + P_2 = 1617\).
Substitute \(P_1\) in terms of \(P_2\) into the sum equation:
\[ \frac{11}{10} P_2 + P_2 = 1617 \]\[ P_2 \left(\frac{11}{10} + 1\right) = 1617 \]\[ P_2 \left(\frac{11+10}{10}\right) = 1617 \]\[ P_2 \left(\frac{21}{10}\right) = 1617 \]Solve for \(P_2\):
\[ P_2 = 1617 \times \frac{10}{21} \]\[ P_2 = 77 \times 10 \]\[ P_2 = 770 \]Now, find \(P_1\) using \(P_1 = 1617 - P_2\):
\[ P_1 = 1617 - 770 \]\[ P_1 = 847 \]Therefore, Shan gave ₹847 to Piyush.
B) ₹847 is the correct amount given to Piyush.
The question asks about the role of the subtropical jet stream in the withdrawal of the monsoon. We need to understand the atmospheric dynamics associated with the jet stream's position and its impact on weather patterns, specifically monsoon withdrawal.
Correct Option: B) Inducing subsidence over land
If \( a^{2} + b^{2} = 168 \), a × b = 27, and a > b, find \(\frac{a-b}{a+b}\).
\(\sqrt{\frac{57}{111}}\)
\(\sqrt{\frac{53}{121}}\)
\(\frac{57}{111}\)
\(\frac{55}{121}\)
To find the value of \(\frac{a-b}{a+b}\), we first need to determine the values of \((a-b)^2\) and \((a+b)^2\). We can use the algebraic identities \((a-b)^2 = a^2 + b^2 - 2ab\) and \((a+b)^2 = a^2 + b^2 + 2ab\). Given \(a^2 + b^2 = 168\) and \(a \times b = 27\), we can substitute these values into the identities.
Calculate \((a-b)^2\):
\[ (a-b)^2 = a^2 + b^2 - 2ab \]Substitute the given values:
\[ (a-b)^2 = 168 - 2(27) \] \[ (a-b)^2 = 168 - 54 \] \[ (a-b)^2 = 114 \]Calculate \((a+b)^2\):
\[ (a+b)^2 = a^2 + b^2 + 2ab \]Substitute the given values:
\[ (a+b)^2 = 168 + 2(27) \] \[ (a+b)^2 = 168 + 54 \] \[ (a+b)^2 = 222 \]Find \(\frac{(a-b)^2}{(a+b)^2}\):
\[ \frac{(a-b)^2}{(a+b)^2} = \frac{114}{222} \]Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 6:
\[ \frac{114 \div 6}{222 \div 6} = \frac{19}{37} \]Find \(\frac{a-b}{a+b}\):
Since we need \(\frac{a-b}{a+b}\), we take the square root of the result from the previous step:
\[ \frac{a-b}{a+b} = \sqrt{\frac{19}{37}} \]However, let's recheck the options and calculations. The options provided are in a different form. Let's re-evaluate the fraction \(\frac{114}{222}\) and compare it with the options. It seems there might be a mistake in the question's options or my simplification. Let's re-examine the options carefully.
Let's re-evaluate the simplification of \(\frac{114}{222}\). Both are divisible by 2:
\[ \frac{114}{222} = \frac{57}{111} \]Now, we can find \(\frac{a-b}{a+b}\):
\[ \frac{a-b}{a+b} = \sqrt{\frac{57}{111}} \]Correct Option: A) \(\sqrt{\frac{57}{111}}\)
The question asks to identify the crop for which 'Aus', 'Aman', and 'Boro' are important varieties, specifically in states like Assam, West Bengal, and Odisha. This requires knowledge of major agricultural crops and their regional varieties in India.
Correct Option: C) Paddy
The question asks about the appointing authority for the Chairman of the National Human Rights Commission (NHRC) of India. This is a factual question related to the constitutional and statutory bodies of India. Understanding the powers of various high offices in India is crucial for answering such questions.
Correct Option: B) The President of India is the appointing authority for the Chairman of the National Human Rights Commission of India, based on the recommendations of a selection committee.
The problem presents a proportion in the format \(a : b :: c : d\). This notation means that the ratio of \(a\) to \(b\) is equal to the ratio of \(c\) to \(d\). In mathematical terms, this can be written as \[\frac{a}{b} = \frac{c}{d}\]. We need to use this property to find the unknown value \(x\).
Understand the proportion: The given expression \(5.1 : 96.9 :: 96.9 : x\) translates to the equation:
\[ \frac{5.1}{96.9} = \frac{96.9}{x} \]Cross-multiply to solve for \(x\): To isolate \(x\), we can cross-multiply the terms:
\[ 5.1 \times x = 96.9 \times 96.9 \]Calculate the product on the right side:
\[ 96.9 \times 96.9 = 9389.61 \]Form the equation with the calculated value:
\[ 5.1x = 9389.61 \]Solve for \(x\): Divide both sides by \(5.1\):
\[ x = \frac{9389.61}{5.1} \]Perform the division:
\[ x = 1841.099... \]Round to one decimal place (as options are given in one decimal place):
Rounding \(1841.099...\) to one decimal place gives \(1841.1\).
D) \(1841.1\) is the correct value of \(x\) obtained from the calculation.
This problem involves calculating work done by individuals and then by a team, followed by one individual completing the remaining work. We will use the concept of work rate (work per day) to solve this.
When A and B work together, their combined one day's work is:
\[ \text{Work by (A+B) in 1 day} = \frac{1}{11} + \frac{1}{33} = \frac{3}{33} + \frac{1}{33} = \frac{4}{33} \]They worked together for 3 days. So, the total work done in these 3 days is:
\[ \text{Work done in 3 days} = 3 \times \frac{4}{33} = \frac{12}{33} = \frac{4}{11} \]The total work is considered as 1 unit. So, the remaining work is:
\[ \text{Remaining work} = 1 - \frac{4}{11} = \frac{11 - 4}{11} = \frac{7}{11} \]A's one day's work is \(\frac{1}{11}\). To find the number of days A will take to complete the remaining \(\frac{7}{11}\) of the work, we divide the remaining work by A's daily work rate:
\[ \text{Days taken by A} = \frac{\text{Remaining work}}{\text{A's 1 day work}} = \frac{7/11}{1/11} = 7 \text{ days} \]Correct Option: A) 7