Trigonometry (Part-6)Total Questions: 5021. cos⁴A - sin⁴A is equal to : [SSC CHSL 03/08/2023 (3rd Shift)](a) 2 cos²A + 1(b) 1 - 2 sin²A(c) 2 sin²A - 1(d) -(2sin²A + 1)Correct Answer: (b) 1 - 2 sin²ASolution:22. Solve it- [SSC CHSL 03/08/2023 (4th Shift)](a) 49/484(b) 357/484(c) 225/484(d) 7/22Correct Answer: (b) 357/484Solution:23. What is the value of the expression. [SSC CHSL 04/08/2023 (4th Shift)](a) secA + cosecA(b) sinA + cosA(c) sinA - cosA(d) secA - cosecACorrect Answer: (a) secA + cosecASolution:24. Simplify the expression. [SSC CHSL 04/08/2023 (4th Shift)](a) 1(b) 0(c) -1(d) 2Correct Answer: (a) 1Solution:25. If secθ + tanθ = √3, then the possitive value of sinθ + cosθ is : [SSC CHSL 04/08/2023 (3rd Shift)](a)(b)(c)(d)Correct Answer: (a)Solution:26. If 2Tan²A + 4Cos⁴A = 3, then the possible value of A is : [SSC CHSL 04/08/2023 (4th Shift)](a) 45º(b) 60º(c) 0º(d) 30ºCorrect Answer: (a) 45ºSolution:27. If a = cotA + cosA and b = cotA - cosA, then find the value of a² - b² - 4 √ab. [SSC CHSL 04/08/2023 (4th Shift)](a) 0(b) -1(c) 1(d) -4Correct Answer: (a) 0Solution:28. The given expression is equal to: [SSC CHSL 07/08/2023 (1st Shift)](a) sin²Φ cos²Φ(b) sin²Φ cot²Φ(c) cot²Φ cos²Φ(d) tan²Φ cos²ΦCorrect Answer: (b) sin²Φ cot²ΦSolution:29. Find the value of cot²B - cosec²B for 0 < B <90º. [SSC CHSL 07/08/2023 (1st Shift)](a) 1(b) 2(c) -1(d) -2Correct Answer: (c) -1Solution:1 + cot²B = cosec²B cot²B - cosec²B = -130. In ΔOPQ, right-angled at P, OP = 7 cm and OQ - PQ = 1cm. Determine the values of sin Q + cos Q . [SSC CHSL 07/08/2023 (2nd Shift)](a) 13/25(b) 21/25(c) 31/25(d) 33/25Correct Answer: (c) 31/25Solution:Submit Quiz« Previous12345Next »