we have to find that non-zero value of k, for which the quadratic equation kx² +1 - 2(k - l)x +x{tex}^2{/tex} = 0 has equal roots. 

Given, {tex}k x ^ { 2 } + 1 - 2 ( k - 1 ) x + x ^ { 2 } = 0{/tex}

{tex}( k + 1 ) x ^ { 2 } - 2 ( k - 1 ) x + 1 = 0{/tex}

For equal roots {tex}D = b ^ { 2 } - 4 a c = 0{/tex}

Here,  a= k + 1,b = -2 (k -1), c = 1

{tex}4 ( k - 1 ) ^ { 2 } - 4 ( k + 1 ) \times 1 = 0{/tex}

or, {tex}4 k ^ { 2 } - 8 k + 4 - 4 k - 4 = 0{/tex}

or, {tex}4 k ^ { 2 } - 12 k = 0{/tex}

or, {tex}4 k ( k - 3 ) = 0{/tex}

{tex}\therefore {/tex} k=3 [k=0 is rejected as it is coefficient of {tex}x^2{/tex}]

By the question we have,

                        (p + 1)x²- 6(p + 1)x + 3(p + 9) = 0. 
Here, a = p + 1, b = -6(p + 1) and c = 3(p + 9)
We know that, D = b² - 4ac
                           = [-6(p + 1)]² - 4(p + 1)[3(p + 9)]
                           = 36(p² + 2p + 1) - (4p + 4)(3p + 27)
                           = 36p² + 72p + 36 - (12p² + 108p + 12p + 108)
                           = 36p² + 72p + 36 - 12p² - 120p - 108
                           = 24p² - 48p - 72

As we know that when quadratic equation has real and equal roots, its discrimnant is equal to zero i.e., 
            D = b² - 4ac = 0

  {tex}\Rightarrow{/tex}      24p² - 48p - 72 = 0
{tex}\Rightarrow{/tex}       p² - 2p - 3 = 0
{tex}\Rightarrow{/tex}      p² - 3p + p - 3 = 0
{tex}\Rightarrow{/tex}      p(p - 3) + 1(p - 3) = 0
{tex}\Rightarrow{/tex}      (p - 3)(p + 1) = 0
{tex}\Rightarrow{/tex}      p - 3 = 0 or p + 1 = 0
{tex}\Rightarrow{/tex}      p = 3 or p = -1

After substituting the values of p in the given equation, the two equations will be of the form
         4x² - 24x + 36 = 0 or 0x² - 0x + 24= 0
Since, 0x² - 0x+24= 0 is not a quadratic equation, therefore we neglect it here.
Now Consider the remaining one i.e.,  4x² - 24x + 36 = 0
{tex}\Rightarrow{/tex}    x² - 6x + 9 = 0
{tex}\Rightarrow{/tex}    x² - 3x - 3x + 9 = 0
{tex}\Rightarrow{/tex}    x(x - 3) - 3(x -3) 0
{tex}\Rightarrow{/tex}     (x - 3)(x - 3) = 0
{tex}\Rightarrow{/tex}    (x - 3)² = 0
{tex}\Rightarrow{/tex}     x - 3 = 0
{tex}\Rightarrow{/tex}     x = 3

Hence the values of x are 3,3.

Let vertices of {tex} \triangle ABC{/tex} be A(x_1, y₁), B(x₂, y₂) and C(x₃, y₃)
By mid-points formula
{tex}\frac{{{x_2} + {x_3}}}{2} = 3 \Rightarrow {x_2} + {x_3} = 6{/tex} ...... (i)
{tex}\frac{{{y_2} + {y_3}}}{2} = 1 \Rightarrow {y_2} + {y_3} = 2{/tex} ...... (ii)
{tex}\frac{{{x_3} + {x_1}}}{2} = 5 \Rightarrow {x_3} + {x_1} = 10{/tex} ..... (iii)
{tex}\frac{{{y_3} + {y_1}}}{2} = 6 \Rightarrow {y_1} + {y_3} = 12{/tex} ..... (iv)
{tex}\frac{{{x_1} + {x_2}}}{2} = - 3 \Rightarrow {x_1} + {x_2} = - 6{/tex} ..... (v)
{tex}\frac{{{y_1} + {y_2}}}{2} = 2 \Rightarrow {y_1} + {y_2} = 4{/tex} ...... (vi)
Adding (i), (iii) and (v)
 
2(x₁ + x₂ + x₃) = 10
{tex} \Rightarrow {/tex} x₁ + x₂ + x₃ = 5 ...... (vii)
Adding (ii),(iv)and (vi)
2(y₁ + y₂ + y₃) = 18
y₁ + y₂ + y₃ = 9 ....... (viii)
Subtracting (i), (iii) and (v) from (vii)
We get, x₁ = -1, x₂ = -5, x₃ = 11
Subtracting (ii), (iv) and (vi) from eq. (viii)
We get, y₁ = 7, y₂ = -3, y₃ = 5

tan A = {tex}=\frac{1}{\cot A}{/tex}
{tex}\Rightarrow{/tex} tan² A = {tex}\frac{1}{\cot ^{2} A}{/tex}
{tex}\Rightarrow{/tex} sec² A - 1 = {tex}\frac{1}{\cot ^{2} A}{/tex}
{tex}\Rightarrow{/tex} sec² A = {tex}\frac{1}{\cot ^{2} A}{/tex}+ 1
{tex}\Rightarrow{/tex} sec² A = {tex}\frac{1+\cot ^{2} A}{\cot ^{2} A}{/tex}
{tex}\Rightarrow{/tex} sec A = {tex}\sqrt{\frac{1+\cot ^{2} A}{\cot ^{2} A}}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{1}{\cos A}=\sqrt{\frac{1+\cot ^{2} A}{\cot ^{2} A}}{/tex}
{tex}\Rightarrow{/tex} cos A = {tex}\sqrt{\frac{\cot ^{2} A}{1+\cot ^{2} A}}{/tex} {tex}=\frac{\cot A}{\sqrt{1+\cot ^{2} A}}{/tex}
{tex}\therefore{/tex} cos A = {tex}\frac{\cot A}{\sqrt{1+\cot ^{2} A}}{/tex}