Exams › JEE Advanced › Maths
In triangle ABC, if b² + c² = 1999 * a², find the value of (cot B + cot C) / cot A.
- 1/999
- 1/1999
- 999
- 1999
Correct answer: 1/999
Solution
Using the formula cot X = (sum of squares of adjacent sides - opposite side squared) / (4 * area): cot B + cot C = [(a²+c²-b²) + (a²+b²-c²)] / (4*Delta) = 2a² / (4*Delta). And cot A = (b²+c²-a²)/(4*Delta). So (cot B + cot C)/cot A = 2a² / (b²+c²-a²). Substituting b²+c² = 1999*a²: denominator = 1999*a² - a² = 1998*a². Ratio = 2a² / (1998*a²) = 1/999.
Related JEE Advanced Maths questions
- In triangle ABC, medians AD and CE have lengths 18 and 27 respectively, and AB = 24. The line CE is extended to meet the circumcircle of triangle ABC at point F. G is the centroid of the triangle. Which of the following statements are correct?
- In a chemistry class, there are 20 students, while a physics class has 30 students. If 10 students are enrolled in both classes, and the two classes are held at separate times, determine the value of k/8, where k represents the total number of students attending either class.
- Given that A represents the divisors of 15, B contains prime numbers less than 10, and C includes even numbers less than 9, how many elements are there in the intersection of (A ∪ C) and B?
- Identify the periodic function among the following:
- The function f(x) = √|x² - 5| x + 6 + √8 + 2|x| - |x|² is defined as a real number for values of x within which range?
- Let f(x) = 4x / (4x² + 2). If the sum of the integrals ∫(1/1997) + ∫(2/1997) +... + ∫(1196/1997) equals 499q, what is the value of q?
⚔️ Practice JEE Advanced Maths free + battle 1v1 →