if 16 cot x = 12, then sin x − cos x sin x + cos x - trigonometric ratios of complementary angles
if 16 cot x = 12, then \[\frac{\sin x - \cos x}{\sin x + \cos x}\]
we are given`16 cot x=12` .we are asked to find the following `(sin x-cos x)/(sin x+cos x)` we know that: `cot x= "base"/"perpendicular" ` → "base"=3 → "perpendicular"=4 → `"hypotenuse"= sqrt(("perpendicular")^2+("base")^2)` → `"hypotenuse"=sqrt(16+9)` →`"hypotenuse"=5` now we have `16 cot x=12` `cot x=12/16` `cot x=3/4`, we know sin x=`"perpendicular"/"hypotenuse" and cos x= "base"/"hypotenuse"` now we find `(sin x- cos x)/(sin z+cos x)` = `(4/5-3/5)/(4/5+3/5)` =`(1/5)/(7/5)` =`1/7`
solution: given that, `sinx+cosx=a`   
 on squaring both sides, we get   
 `" "(sinx+cosx)^(2)=(a)^(2)`   
 `rarr" "sin^(2)x+cos^(2)x+2sinxcosx=a^(2)`   
 `rarr" "sinx*cosx=(1)/(2)(a^(2)-1)`   
 `(i) sin^(6)x+cos^(6)x=(sin^(2)x)^(3)+(cos^(2)x)^(3)`   
 `" "=(sin^(2)x+cos^(2)x)(sin^(4)x-sin^(2)xcos^(2)x+cos^(4)x)` ltbrgt `" "=sin^(4)x+cos^(4)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=1-2*(1)/(4)(a^(2)-1)^(2)-(1)/(4)(a^(2)-1)^(2)=(1)/(4)[4-3(a^(2)-1)^(2)]`   
 `(ii) abs(sinx-cosx)=sqrt((sinx-cosx)^(2))`   
 `" "=sqrt(sin^(2)x+cos^(2)x-2sinxcosx)`   
 `" "=sqrt(1-2(1)/(2)(a^(2)-1))=sqrt(1-a^(2)+1)=sqrt(2-a^(2))` doing integration yields-then finally f(x)=-x cos x + sin xto calculate derivative of the above function f(x) we have to apply formula of derivation of products of two functions-f(x) =-x cos(x)+sin(x)f’(x) =-[x(-sin x)+cos x.1]+cos x ;{by formula d/dx (f.g)=f.g’+g.f’}f’(x) =-(-x sin x) - cos x + cos xf’(x) = x.sin(x) dear student,
1+ tan2x= 1+sin2x/cos2x
find common denominator which is cos2x
cos2x/cos2x +
sin2x/cos2x
(cos2x + sin2x)/cos2x
1+cot2x = 1+ cos2x/sin2x
common denominator is sin2x
sin2x/sin2x +
cos2x/sin2x
( sin2x + cos2x)/sin2x
1/(cosx2+ sin2x)/cos2x + 1/(
sin2x + cos2x)/sin2x
multiply the numerator and denominator by the reciprocal of the
bottom fractions to get
cos2x/(cosx2+ sin2x )+
sin2x/( cosx2+ sin2x)
since they now have a common denominator you can add the
tips
(cosx2+ sin2x)/(cosx2+
sin2x)
1/1 = 1

regards given;y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2⇒ sin x = cos xrequired area= (√2 − 1) sq.units 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。 let a, b, c be the sides of any triangle abc. then by applying the sine rule, we get⇒c = k sin csimilarly, b = k sin band a = k sin ahere we will consider rhs, so we getrhs = (b2 – c2) sin asubstituting corresponding values in the above equation, we get⇒ = [( k sin b)2 - ( k sin c)2] sin a⇒ = k2(sin2 b – sin2 c )sin a……….(ii)but,substituting the above values in equation (ii), we get⇒ = k2(sin(b + c) sin(b - c)) sin abut a + b + c = π ⇒ b + c = π –a, so the above equation becomes,⇒ = k2(sin(π –a) sin(b - c))sin abut sin (π - θ) = sin θ⇒ = k2(sin(a) sin(b - c))sin arearranging the above equation we get⇒ = (k sin(a))( sin(b - c))(k sin a)from sine rule, a = k sin a, so the above equation becomes,⇒ = a2 sin(b - c) = rhshence proved the value of is we need, principal value of cos-1 x is [0,π]also, cos (2nπ + θ) = cos θ for all n є nso, switch x and y, then solve again for y:

x = 3y+12
x-12 = 3y
(x-12)/3 = y

or,

y = x/3 - 4 the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is