Question
If 9 sinx + 40 cosx = 41 then find tanx.
Solution
9 sinx + 40 cosx = 41 or (9/41) sinx + (40/41) cosx = 1 ---(1) Now comparing equation with sin2 x + cos2 x = 1 ; We can say that, sinx = 9/41 and cosx =40/41 Hence tanx = sinx/cosx = (9/41)/(40/41) = 9/40 ` `
((67)32 × (67)-18 / ? = (67)βΈ
What will come in the place of question mark (?) in the given expression?
?% of (112 X 3 + 164) + 75 = 2 X 140 + 35
20% of 450 - 15% of 400 = 25% of ?
β121 + β961β β289 =?2
45% of 1020 + ?% of 960 = 747

(γ(0.4)γ^(1/3)Β Γ γ(1/64)γ^(1/4)Β Γ γ16γ^(1/6)Β Γ γ(0.256)γ^(2/3))/(γ(0.16)γ^(2/3)Β Γ 4^(-1/2)Β Γγ1024γ^(-1/4) ) = ?
(512) (2/3) Γ β64 Γ· (512) (1/3) = (64) (?/2) Γ· (2)6Β
1231 + 1312 + 2113 β 3211 = ?
Simplify the following expression:-