#### Explain Solution R. D. Sharma Class 12 Chapter deteminants Exercise 5 .2 Question 2 Sub Question 17 maths Textbook Solution.

Answer: $\left|\begin{array}{lll} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|=0$where A,B,C are the angles of ΔABC

Hint: We will try to convert some elements of determinant zero or 1

Given:$\left|\begin{array}{ccc} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|$

Solution: $\left|\begin{array}{ccc} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3}\\ &=\left|\begin{array}{ccc} \sin ^{2} A-\sin ^{2} B & \cot A-\cot B & 0 \\ \sin ^{2} B-\sin ^{2} C & \cot B-\cot C & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|\\ &\because \sin ^{2} A-\sin ^{2} B=\sin (A+B) \cdot \sin (A-B) \backslash\\ &\because A+B+C=180^{\circ}\\ &A-B=180-C\\ &\sin (180-C)=\sin C\\ &\sin (180-A)=\sin A \end{aligned}

\begin{aligned} &\text { Now, }\\ &\cot A-\cot B\\ &\frac{\cos A}{\sin A}-\frac{\cos B}{\sin B}\\ &=\frac{\sin B \cos A-\cos B \sin A}{\sin A \sin B} \Rightarrow \frac{\sin (B-A)}{\sin A \sin B}\\ &=\frac{\sin \{-(A-B)\}}{\sin A \sin B}=\frac{-\sin (A-B)}{\sin A \sin B} \end{aligned}

\begin{aligned} &\text { Now from (1) }\\ &=\left|\begin{array}{ccc} \sin (A+B) \sin (A-B) & \frac{\sin (B-A)}{\sin A \sin B} & 0 \\ \sin (B+C) \sin (B-C) & \frac{\sin (C-B)}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right| \end{aligned}

\begin{aligned} &=\left|\begin{array}{ccc} \sin (180-C) \sin (A-B) & \frac{-\sin (A-B)}{\sin A \sin B} & 0 \\ \sin (180-A) \sin (B-C) & \frac{-\sin (B-C)}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right| \\ &=\left|\begin{array}{ccc} \sin (A) \sin (B-C) & \frac{-\sin (B-C)}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking } \sin (A-B) \text { common from } \mathrm{R}_{1} \text { and } \sin (B-C) \text { from } \mathrm{R}_{2}\\ &=\sin (A-B) \cdot \sin (B-C)\left|\begin{array}{lcc} \sin C & \frac{-1}{\sin A \sin B} & 0 \\ \sin A & \frac{-1}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{3}\\ &=\sin (A-B) \cdot \sin (B-C)\left[0\left|\begin{array}{lc} \sin A & \frac{-1}{\sin B \sin C} \\ \sin ^{2} C & \cot C \end{array}\right|-0\left|\begin{array}{lc} \sin C & \frac{-1}{\sin A \sin B} \\ \sin ^{2} C & \cot C \end{array}\right|+1\left|\begin{array}{ll} \sin C & \frac{-1}{\sin A \sin B} \\ \sin A & \frac{-1}{\sin B \sin C} \end{array}\right|\right]\\ &=\sin (A-B) \cdot \sin (B-C)\left[0-0+1\left(\frac{-\sin C}{\sin B \sin C}+\frac{\sin A}{\sin A \sin B}\right)\right]\\ &=\sin (A-B) \cdot \sin (B-C)\left[\frac{-1}{\sin B}+\frac{1}{\sin B}\right]\\ &=\sin (A-B) \cdot \sin (B-C) \times 0\\ &=0 \end{aligned}

Hence $\left|\begin{array}{lll} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|=0$