Question
If secθ + tanθ = 3+√10, then the value
of sinθ+cosθ isSolution
secθ + tanθ = 3+√10 ............... (i) (sec²θ - tan²θ) = 1 (secθ + tanθ) (secθ - tanθ) = 1 (3+√10) (secθ - tanθ) = 1 secθ - tanθ = 1/(3+√10) = 1/(√10+3) secθ - tanθ = √10-3 ............ (ii) From (i) and (ii) we get, 2secθ = 2√10 secθ = √10 cosθ = 1/√10 sin²θ + cos²θ = 1 sin²θ + 1/10 = 1 sin²θ = 1- 1/10 = 9/10 sinθ = 3/√10 sinθ + cosθ = 3/√10 + 1/√10 = 4/√10 = 4/√10 × √10/√10 = (2√10)/5
15.99% of 549.99 Γ· 11.17 = ? Γ· 20.15
74.91% of 639.95 β 599.98% of 45 + 119.987 = ?
(4.88 Γ 5.76)2 - ?2 = 39.89 Γ 19.86
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exactvalue.)
(1800.23 Γ· 29.98) + (816.32 Γ· 23.9) + 1634.11 = ?
1449.98 Γ· 50.48 Γ 10.12 = ? Γ 2.16
36.05 Γ 5.02 + 12.052 = ? + 9.09 Γ 4.04Β
(31.9)3 + (34.021)Β² - (16.11)3 - (42.98)Β² = ?