how to find value of tan inverse 1 (12/35) someone said 19 degree how - maths - introduction to trigonometry
how to find value of tan inverse 1 (12/35) someone said 19 degree
how
dear student
value of tan-11235=18 92 by calculator
`sin^2 30° cos^2 45 ° + 4 tan^2 30° + 1/2 sin^2 90° − 2 cos^2 90° + 1/24 cos^2 0°` ....(i) by trigonometric ratios we have `sin 30^@ = 1/2` `cos 45^@ = 1/sqrt2` `tan 30^2 = 1/sqrt3` `sin 90^@ = 1 cos 90^@ = 0 cos 0^@ = 1` by substituting above values in (i), we get `[1/2]^2 . [1/sqrt2]^2 + 4[1/sqrt3]^2 + 1/2[1]^2 - 2[0]^2 + 1/24 [1]^2` `1/4.1/2 + 4/ 1/3 + 1/2 - 0 + 1/24` `1/8 + 4/3 + 1/2 + 1/24 = 48/24 = 2` tan−1{2cos(2sin−121​)}=tan−1{2cos(2×6π​)}=tan−1{2cos(3π​)}=tan−1{2×21​}=tan−1{1}=4π​ you can find the radius in the following way. 
write the equation of the line that is perpendicular to the given tangent line. this perpendicular goes through the stated center, and intersects the tangent; the perpendicular's slope is the negative reciprocal of the stated tangent line.
find the intersection (point t) of the tangent and radius line--then use the distance formula to find the radius, which is a segment from the stated center to the point t. given = let, tan-1x3 = z⇒ ⇒ so,where c is the integrating constant. the formulas for converting between degree celsius and degree fahrenheit are: °f = (°c * 9/5) + 32 °c = (°f - 32) * 5/9 to find the temperature when both are equal, we use an old algebra trick and just set of = oc and solve one of the equations. °c = (°c * 9/5) + 32 °c - (°c * 9/5) = 32 - 4/5 * °c = 32 °c = - 32 * 5/4 °c = - 40 °f = (°f * 9/5) + 32 °f - (°f * 9/5) = 32 - 4/5 * °f = 32 °f = - 32 * 5/4 °f = - 40 so the temperature when both the celsius and fahrenheit scales are the same is - 40 degrees. sin(theta) = d, 
cos(theta) = a, 
tan(theta) = h, 
csc(theta) = f, 
sec(theta) = c, 
cot(theta) = g. sin(α β) = sinαcosβ cosαsinβ 
sin(α-β) = sinαcosβ - cosαsinβ


cos(α β) = cosαcosβ - sinαsinβ
cos(α-β) = cosαcosβ sinαsinβ


tan(α β) = tanα tanβ / 1-tanα tanβ
tan(α-β) = tanα-tanβ / 1 tanα tanβ


sin2θ = 2sinθcosθ = 2tanθ /1 tanθtanθ


cos2θ = cos平方θ−sin平方θ = 2cos平方θ−1 = 1−2sin平方θ


△abc=海龍公式 √s(s-a)(s-b)(s-c) 
[s=a b c/2 ，就是周長/2 ]



△abc=1/2 ａ.b*sinc = axbxc / 4r = rxs 

 [r=外接圓半徑，r=內接圓半徑]




應該是這樣吧!希望對你有幫助。 setup the kc expression for the reaction, recall that kc expression will appear as:kc expression for the problem:solving for [n2]:note that ice table is not necessary for this calculation answer is 0.016 m [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2] let, y = (1 – 2 tan x)(5 + 4 sin x)we have to find dy/dxas we can observe that y is a product of two functions say u and v where,u = (1 – 2tan x) and v = (5 + 4sin x)∴ y = uvas we know that to find the derivative of product of two function we apply product rule of differentiation.by product rule, we have –…equation 1as, u = (1 – 2tan x)∴ ⇒ ⇒ …..equation 2 {∵ }as, v = 5 + 4sin x⇒ ⇒ …equation 3 {∵ }∴ from equation 1, we can find dy/dx∴ using equation 2 & 3, we get –⇒ ⇒ ∵ sin x = tan x cos x , so we get –⇒ hence,