Two consecutive terms in the expansion of (3+2x)^74 whose coefficients are equal are ….?

cos-1 ([x2-1]/[x2+1]) + tan-1[(2x)/(x2-1)] =2π/3, then x=?

Q. cos-1 ([x²-1]/[x²+1]) + tan-1[(2x)/(x²-1)] =2π/3, then x=?

Solution: Let,

cos-1 ([x²-1]/[x²+1]) =θ-------------------1

→cos θ= (x²-1)/(x²+1)

→cos² θ=(x²-1)²/(x²+1)²

→1- cos²θ=1-(x²-1)²/(x²+1)²

              =[(x²+1)²-(x²-1)²]/ (x²+1)²

              = [2x²X 2]/(x²+1)²

→Sin²θ= 4x²/(x²+1)²

→sin θ=[2x]/(x²+1)

→sin θ/ cos θ=2x/(1-x²) =tan θ

→θ=tan^-12x/(1-x²) --------------------------2

From 1 and 2,

→cos-1 ([x²-1]/[x²+1]) + tan-1[(2x)/(x²-1)] =2π/3

→2 θ= 2π/3

→θ= π/3

so, cos θ= (x²-1)/(x²+1) =cos π/3

 → (x²-1)/(x²+1) =1/2

→X²+1=2-2x²

→3x²-1=0

→ (√3x+1) (√3x-1) =0

So, x=±1/√3(Ans)

sin-1 x+sin-1 y+sin-1 z=π then x2+y2+z2+2xyz=?

sin-1 x+sin-1 y+sin-1 z=π then x2+y2+z2+2xyz=?

Solution:

Let A = sin^-1(x), B= sin^-1( y) , C = sin^-1( z ) ,

=> X=Sin A, Y=Sin B, Z=Sin Z 

then given A+B+C = pi/2,

=>A+B= pi/2-C

Apply cos on either side  

=>Cos(A+B) = Cos (pi/2 - C) = sin C

=>Cos A. Cos B - sin A. sin B = sin C

=>Cos A. cos B = sin A. sin B + sin C

=>√ (1 - x²) √(1 - y²) = xy+z [squaring both  sides]

=>1 - x² - y² + x² y² = x² y² + z² + 2xyz

=> x^2 +y^2 +z^2 +2xyz = 1 (Answer)

If the distance between the parallel lines x+2y+3=0 and x+2y+k=0 is √5, then the value of k is?

Q.If the distance between the parallel lines x+2y+3=0 and x+2y+k=0 is √5, then the value of k is?
Solution:

 As shown in the graph the given two lines, their slopes, the perpewndicular distance between two lines and the coordinates of various points which would immideatly help solving this question given,
              The perpendicular distance between two lines(AB)=√5

               The value of tanθ= 1/2 = slope of two lines
In the triangle ABC,      tanθ=1/2=[√5]/BC, Therefore, BC=2√5
                                       AC^2    =AB^2+BC^2
                                       (k-3)^2 =(√5)^2+(2√5)^2
                Therefore,                k  = 8(ans)


                                                                                                                      -Mimansa