prove the following identitiessecθ (1 – sinθ)(secθ + tanθ) = 1
consider lhs,lhs = secθ (1 – sinθ) (secθ + tanθ)⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ - ) (secθ + tanθ)we know that = tanθ.⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ – tanθ) (secθ + tanθ)we know that (a + b) (a – b) = a2 – b2.⇒ secθ (1 – sinθ) (secθ + tanθ) = sec2θ – tan2θwe know that sec2θ – tan2θ = 1.∴ secθ (1 – sinθ) (secθ + tanθ) = 1 = rhshence proved.
sin(theta) = d, 
cos(theta) = a, 
tan(theta) = h, 
csc(theta) = f, 
sec(theta) = c, 
cot(theta) = g. let i = `(1)/(sinx*(3 + 2cosx))*dx` = `int sinx/(sin^2x*(3 + 2cosx))*dx` = `int sinx/((1 - cos^2x)(3 + 2cosx))*dx` = `int sinx/((1 - cosx)(1 + cosx)(3 + 2cosx))*dx` put cos x = t∴ – sinx.dx = dt∴ sinx.dx = – dt ∴ i = `int (1)/((1 - t)(1 + t)(3 + 2t))*(-dt)` = `int (-1)/((1 - t)(1 + t)(3 + 2t))*dt` let `(-1)/((1 - t)(1 + t)(3 + 2t)) = "a"/(1 - t) + "b"/(1 + t) + "c"/(3 + 2t)` ∴ – 1 = a(1 + t)(3 + 2t) + b(1 - t)(3 + 2t) + c(1 - t)(1 + t)put 1 – t = 0, i.e. t = 1, we get– 1 = a(2)(5) + b(0)(5) + c(0)(2) ∴ – 1 = 10a ∴ a = `(-1)/(10)`put 1 + t = 0, i.e. t = – 1, we get– 1 = a(0)(1) + b(2)(1) + c(2)(0) ∴ – 1 = 2b ∴ b = `-(1)/(2)` put 3 + 2t = 0, i.e. t = `-(3)/(2)`, we get – 1 = `"a"(-1/2)(0) + "b"(5/2)(0) + "c"(5/2)(-1/2)` ∴ – 1 = `-(5)/(4)"c"` ∴ c = `(4)/(5)` ∴ `(-1)/((1 - t)(1 + t)(3 + 2t)) = (((-1)/(10)))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)` ∴ i = `int [(((-1)/10))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)]*dt` = `-(1)/(10) int 1/(1 - t)*dt - (1)/(2) int 1/(1 + t)*dt + (4)/(5) int 1/(3 + 2t)*dt` = `-(1)/(10) (log|1 - t|)/(-1) - (1)/(2) log | 1 + t| + 4/5 (log|3 + 2t|)/(2) + c` = `(1)/(10)log|1 - cosx| - (1)/(2)log|1 + cosx| + (2)/(5)log|3 + 2cos| + c`. tan−1{2cos(2sin−121​)}=tan−1{2cos(2×6π​)}=tan−1{2cos(3π​)}=tan−1{2×21​}=tan−1{1}=4π​ 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 to prove 

then taking l.h.s.

we have 

                    





                                  

 

  

=  r.h.s.

hence proved. tan a/2 is equal to 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。 answer: k=7 the sum of a pmf over its entire domain must equal 1.
so cς(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = 1.
 
ς(i=1 to ∞)[1/n^i] = 1/(n-1). so
 
ς(y=1 to ∞)[(1/3^y)] = 1/(3-1) = 1/2 and
ς(x=1 to ∞)[(1/4^x)] = 1/(4-1) =1/3.
 
so, σ(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = (1/2)*(1/3) = 1/6.
 
to make the sum = 1, then c=6. the first one.