Question

12 class maths determinants please help me out​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\: \begin{gathered}\sf \left | \begin{array}{ccc}cosec \theta \: &1&0\\1&2cosec \theta& 1\\0& 1&2cosec \theta\end{array}\right | \end{gathered}$

$\rm :\longmapsto\:OP \: R_2 \: \to \: cosec \theta \: R_2$

$\rm \: = \dfrac{1}{cosec \theta} \: \begin{gathered}\sf \left | \begin{array}{ccc}cosec \theta \: &1&0\\cosec \theta&2cosec^{2} \theta& cosec \theta\\0& 1&2cosec \theta\end{array}\right | \end{gathered}$

$\rm :\longmapsto\:OP \: R_2 \: \to \: \: R_2 - R_1$

$\rm \: = \dfrac{1}{cosec \theta} \: \begin{gathered}\sf \left | \begin{array}{ccc}cosec \theta \: &1&0\\0&2cosec^{2} \theta - 1& cosec \theta\\0& 1&2cosec \theta\end{array}\right | \end{gathered}$

Now, expand along Column 1, we get

$\rm \: = \dfrac{1}{cosec \theta} \times cosec \theta (4{cosec}^{3}\theta \: - 2cosec \theta - cosec \theta)$

$\rm \: = 4{cosec}^{3}\theta \: - 3cosec \theta$

$\rm :\longmapsto\: \begin{gathered}\sf \left | \begin{array}{ccc}cosec \theta \: &1&0\\1&2cosec \theta& 1\\0& 1&2cosec \theta\end{array}\right | \end{gathered} = 4 {cosec}^{3}\theta - cosec\theta - - (1)$

Now,

Consider,

$\rm :\longmapsto\:tan\dfrac{\theta}{2} + cot\dfrac{\theta}{2}$

$\rm \:  =  \:  \:\dfrac{sin\dfrac{\theta}{2}}{cos\dfrac{\theta}{2}} + \dfrac{cos\dfrac{\theta}{2}}{sin\dfrac{\theta}{2}}$

$\rm \:  =  \:  \:\dfrac{ {sin}^{2}\dfrac{\theta}{2} + {cos}^{2}\dfrac{\theta}{2}}{sin\dfrac{\theta}{2} \: cos\dfrac{\theta}{2}}$

$\rm \:  =  \:  \:\dfrac{ 1}{sin\dfrac{\theta}{2} \: cos\dfrac{\theta}{2}}$

$\rm \:  =  \:  \:\dfrac{2}{2 \: sin\dfrac{\theta}{2} \: cos\dfrac{\theta}{2}}$

$\rm \:  =  \:  \:\dfrac{2}{sin\theta \:}$

$\rm \:  =  \:  \:2cosec\theta \:$

Now,

Consider RHS

$\rm :\longmapsto\:\dfrac{1}{2}\bigg( {tan}^{3}\dfrac{\theta}{2} + {cot}^{3}\dfrac{\theta}{2} \bigg)$

We know that

$\rm :\longmapsto\: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)$

So,

Consider,

$\rm :\longmapsto\:{tan}^{3}\dfrac{\theta}{2} + {cot}^{3}\dfrac{\theta}{2}$

$\rm \:  =  \:  \: {\bigg(tan\dfrac{\theta}{2} + cot\dfrac{\theta}{2}\bigg) }^{3} -3tan\dfrac{\theta}{2}cot\dfrac{\theta}{2} \bigg(tan\dfrac{\theta}{2} + cot\dfrac{\theta}{2}\bigg)$

$\rm \:  =  \:  \: {\bigg(tan\dfrac{\theta}{2} + cot\dfrac{\theta}{2}\bigg) }^{3} -3 \bigg(tan\dfrac{\theta}{2} + cot\dfrac{\theta}{2}\bigg)$

$\rm \:  =  \:  \: {(2cosec\theta \:)}^{3} - 3(2cosec\theta \:)$

$\rm \:  =  \:  \:8 {cosec}^{3}\theta \: - 6 {cosec}\theta \:$

$\bf\implies \: {tan}^{3}\dfrac{\theta}{2} + {cot}^{3}\dfrac{\theta}{2}=\:8 {cosec}^{3}\theta \: - 6 {cosec}\theta \:$

$\bf\implies \: {tan}^{3}\dfrac{\theta}{2} + {cot}^{3}\dfrac{\theta}{2}=2(4{cosec}^{3}\theta- 3{cosec}\theta)$

$\bf\implies \:\dfrac{1}{2}\bigg({tan}^{3}\dfrac{\theta}{2} + {cot}^{3}\dfrac{\theta}{2}\bigg)=4{cosec}^{3}\theta- 3{cosec}\theta - - (2)$

Hence, From equation (1) and equation (2), we concluded that

$\rm :\longmapsto\: \begin{gathered}\bf \left | \begin{array}{ccc}cosec \theta \: &1&0\\1&2cosec \theta& 1\\0& 1&2cosec \theta\end{array}\right | \end{gathered} = \dfrac{1}{2}\bigg( {tan}^{3}\dfrac{\theta}{2} + {cot}^{3}\dfrac{\theta}{2}\bigg)$

Hence, Proved