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A triangle is inscribed in a circle. the vertices of the triangle divided the circumference of the circle into three area of length 6,8,10 units then the area of triangle is equal to... (a) ((64(√3)((√3)+1))/π^2 ) (c) ((36(√3)((√3)−1))/π^2 ) (b) ((72(√3)((√3)+1))/π^2 ) (d) ((36(√3)((√3)+1))/π^2 ) |
Prove that ∫^( x) _0 (t/(e^t −1)) dt = Σ_(n=1) ^(+∞) (((1−e^(−x) )^n )/n^2 ) |
lim_(x→0) (((x+y)sec (x+y)−ysec y)/x)=? |
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x;y;z>0 ; x+y+z=3 proof (x^3 +2)(y^3 +2)(z^3 +2)≥3^3 |
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n ∈ N^∗ and k ∈ N^∗ . Given 0≤k≤n−1. Show that (1/n)ln(1+(k/n))≤∫_(1+(k/n)) ^(1+((k+1)/n)) lnx dn≤(1/n)ln(1+((k+1)/n)) |
e^((((ζ(2))/2)−((ζ(3))/3)+((ζ(4))/4)−((ζ(5))/5)+...)) =? |
Prove it (√(a+(√b)))=(√(((a+(√(a^2 +b)))/2)+))(√((a−(√(a^2 −b)))/2)) |
If three vector a^→ , b^→ and c^→ are such that a^→ ≠ 0 and a^→ ×b^→ = 2(a^→ ×c^→ ) ,∣a^→ ∣ = ∣c^→ ∣ = 1 , ∣b^→ ∣ = 4 and the angle between b^→ and c^→ is cos^(−1) ((1/4)), then b^→ −2c^→ = λ a^→ , where λ =? |
tan^2 1°+tan^2 2°+tan^2 3°+...+tan^2 89°=((15931)/3) ??? |
........ nice ....... calculus....... simplify: 𝛗(x):= sin((x/2))(1+2Σ_(m=1) ^n cos(mx)) |
sin^(−1) (sin x)=x sin^(−1) (cos x)=? sin^(−1) (tan x)=? |
If lim_(x→0) (cos x + a sin bx)^(1/x) = e^2 { ((a=?)),((b=?)) :} |
(f(x))^2 . f(((1−x)/(1+x))) = 64x , ∀x∈D ⇒ f(x) =? |
a;b;c∈R ; ∀∣x∣≤1 ∣ax^4 +bx^2 +c∣(√(1−x^2 ))≤1 ; proof ∣a∣≤16 |
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Find all 2×2 matrices A with A^3 −3A^2 = (((−2 −2)),((−2 −2)) ) . |
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∫_0 ^∞ ((1/( (√(1+x))))−(1/( (√(1+x^2 )))))(dx/x)=log(2) |
Pg 728 Pg 729 Pg 730 Pg 731 Pg 732 Pg 733 Pg 734 Pg 735 Pg 736 Pg 737 |