#### Provide Solution For  R. D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 2 Sub Question 12 Maths Textbook Solution.

Answer:$\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|=0$

Hint: We will try to do any two column or row equal

Given:$\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|$

Solution:

\begin{aligned} &\text { First }(a+b)^{2}-(a-b)^{2} \\ &=\left(a^{2}+2 a b+b^{2}\right)-\left(a^{2}-2 a b+b^{2}\right) \\ &=a^{2}+2 a b+b^{2}-a^{2}+2 a b-b^{2} \\ &=4 a b \end{aligned}

$\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \\ &\qquad \begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2}-\left(2^{x}-2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2}-\left(3^{x}-3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2}-\left(4^{x}-4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array} \mid \\ &=\left|\begin{array}{lll} 4 \times 2^{x} \times 2^{-x} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 \times 3^{x} \times 3^{-x} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 \times 4^{x} \times 4^{-x} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| \end{aligned}

\begin{aligned} &\left\{\begin{array}{l} (a+b)^{2}-(a-b)^{2}=4 a b \\ a^{m} \times a^{n}=a^{m+n} \end{array}\right\} \\ &=\left|\begin{array}{lll} 4 \times 2^{x+(-x)} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 \times 3^{x+(-x)} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 \times 4^{x+(-x)} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| \\ &=\left|\begin{array}{lll} 4 \times 2^{0} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 \times 3^{0} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 \times 4^{0} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| \end{aligned}

\begin{aligned} &=\left|\begin{array}{lll} 4 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Taking } 4 \text { common from } \mathrm{C}_{1}\\ &=4\left|\begin{array}{lll} 1 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 1 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 1 & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| \end{aligned}

If any two rows or columns of a determinant are identical.

The value of the determinant is zero

\begin{aligned} &=4 \times 0 \quad\left(\because C_{1}=C_{3}\right) \\ &=0 \end{aligned}

Hence $\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|=0$