Question

Please try to answer IT IS SO URGENT....

remained er and factor theorem :
given that x+2andx-3are factors of x^3+ax+b.calculate the values of a and b.also find the remaining factor

use the remainder theorem to factorise the expression 2x^3+9x^2+7x-6.hence solve the equation 2x^3+9x^2+7x-6=0

trigonometry:
cosecA+1/cosec A-1=(secA+tanA)^2

cotA-tanA=2cos^2A-1/sinAcosA

√1+cosA/1-cosA +√1-cosa/1+cosa=2coscA

cosa/cosecA+1+cosa/cosecA-1=2tanA

1+cotA/cosa+1+tanA/sins=2(secA+cosecA)

√1+sins/1-sinA-√1-sinA/1+sins=2tanA

Answer 1

1.
Solution:

Given: (x+2) and (x-3) are factors of equation x^3+ax+b,.
To find: values of a and b (a=?, b=?)

 If, (x+2) is a factor,

   x+2=0
=>x=-2
By using remainder theorem we get,

 p(x)=x^3+ax+b

 p(-2)=(-2)^3+a(-2)+b

    0  =(-8)-2a+b
    0  =-8-2a+b
   
    -2a+b=8 ....(i)

If (x-3) is a factor,

     x-3=0
=>x=3

By using remainder theorem we get,

p(x)=x^3+ax+b
p(3)=(3)^3+a(3)+b
   0 =27+3a+b

   3a+b=-27...(ii)

Subtracting (i) from (ii) we get,
 
                 (3a+b) - (-2a+b) = (-27)-(8)
                         3a+b+2a-b = -27-8
                                         5a=-35
                                           a=-7

Substituting value of a in (ii) we get,

                                     3a+b=-27
                                  3(-7)+b=-27
                                    -21+b=-27
                                            b=-27+21
                                            b=-6

2.

Solution:
 
Given: p(x)= 2x^3+9x^2+7x-6

By Using Factorization Method;

Let x=1/2
            p(x)=2x^3+9x^2+7x-6
            p(1/2)=2(1/2)^3+9(1/2)^2+7(1/2)-6
            p(1/2)=2(1/8)+9(1/4)+(7/2)-6
            p(1/2)=1/4+9/4+7/2-6
            p(1/2)=10/4+14/4-24/4
            p(1/2)=0......(x-1/2) x 2 =>(2x-1) is a factor

Let x=-2
            p(x)=2x^3+9x^2+7x-6
            p(-2)=2(-2)^3+9(-2)^2+7(-2)-6
            p(-2)=2(-8)+9(4)+(-14)-6
            p(-2)=-16+36-14-6
            p(-2)=0.........(x+2) is a factor

Let x= -3
            p(x)=2x^3+9x^2+7x-6
            p(-3)=2(-3)^3+9(-3)^2+7(-3)-6
            p(-3)=2(-27)+9(9)-21-6
            p(-3)=-54+81-27
            p(-3)=0........(x+3) is a factor

            p(x)=2x^3+9x^2+7x-6=(2x-1)(x-2)(x+3)

3.
Solution:

To prove: cosec A + 1/ cosec A - 1=(sec A + tan A)^ 2

proof: 
 LHS=>  cosec A + 1/ cosec A - 1
         =>(cosecA+1/cosecA-1)x(cosec A+1/ cosecA+1)  (by taking conjugate)
         =>[(cosec A + 1)^ 2/ (cosec^ 2 A - 1)]
         =>[(cosec^2+1+2cosecA)/cot^2A]             (cosec^2A-1=cot^2A)
         =>[(cot^2A+2+2cosecA)/(1/tan^2A)
         =>[(1/tan^2A+2(1+cosecA))/tan^2A]x(tan^2A)/1
         =>[{1+2tan^2A(1+cosecA)}]
         =>[1+2tan^2A+2tan^2AcosecA]
         =>(1+2sin^2A/cos^2A+2sin^2AcosecA/cos^2A)
         =>(1+{2sin^2A}{1/cos^2A}+{2sin^2A(1/ sinA)}{1/cos^2A})
         =>(1+{2sin^2A}{sec^2A}+{2sinA}{sec^2A})        
         =>(1+2sin^2Asec^2A+2sinAsec^2A)
        =>(secA+tanA)^2=RHS
                        ∴Hence proved
same way,you need to prove the remaining parts