Question
If 2cos²A + 3sin²A = 13/5, then find the value of (sec²A
- 1)Solution
2cos²A + 3sin²A = 13/5
Or, -1cos²A + 3(cos²A + sin²A) = 13/5
Or, -cos²A + 3 = 13/5
Or, -cos²A = 13/5 - 3 = -2/5 => cos²A = 2/5
So, sec²A = 1 / (2/5) = 5/2
Required value = 5/2 - 1 = 3/2
(1440.13 ÷ 31.77) × (√168.69 + 16) - (24.87% of 719.99) = ?
`sqrt(224.991)` + (3.81008)2 = ? `-:` `1/6.0001`
` `
The greatest number that will divide 398,436, and 542 leaving 7, 11, and 15 as remainders, respectively, is:
1254.04 – 440.18 + 399.98 ÷ 10.06 = ?
(1011 × 4.465 + 1012 × 0.3535) ÷ (160 × 10⁵) = 10? ÷ 2
...A wire is bent to form a square whose sides are of 132 cm. If the same wire is bent to form a circle, then find the area (in cm2) of the circ...
(84.92 + 235.17) ÷ (15.93 × 3.89) = ? ÷ 21.02
(?)2 + 8.113 = 28.92 – 73.03
6106.11 ÷ √? × 55.9 = 3976.21
139.88% of 119.89 + 1451.89 ÷ 6.01 - √196.01 = ? ÷ 3.01 + 215.98
Relevant for Exams:
