Question

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$[1 + \frac{1}{ \tan^{2}θ }] [1 + \frac{1}{ \cot^{2}θ } ] = \frac{1}{ \sin^{2} θ - \sin^{4}θ }$
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Answer 1

Step-by-step explanation:

We have,

$\left \{ 1 + \dfrac{1}{ \tan^{2} ( \theta) } \right \} \left \{ 1 + \dfrac{1}{ \cot^{2} ( \theta) } \right \}$

$= \left \{ 1 + \cot^{2} ( \theta) \right \} \left \{ 1 +\tan^{2} ( \theta) \right \}$

$= \cosec^{2} ( \theta) \cdot \sec^{2} ( \theta)$

$= \dfrac{1}{ \cos^{2} ( \theta) \cdot \sin^{2} ( \theta)}$

$= \dfrac{1}{ \sin^{2} ( \theta) \cdot(1 - \sin^{2} ( \theta))}$

$= \dfrac{1}{ \sin^{2} ( \theta) - \sin^{4} ( \theta)}$

Answer 2

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

Consider LHS

$\rm :\longmapsto\:\bigg[1 + \dfrac{1}{ {tan}^{2}\theta } \bigg]\bigg[1 + \dfrac{1}{ {cot}^{2} \theta} \bigg]$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \frac{1}{tanx} = cotx}}} \: and \: \red{\boxed{\tt{ \frac{1}{cotx} = tanx}}}$

So, using this, we get

$\rm \:  =  \: (1 + {cot}^{2}\theta)(1 + {tan}^{2}\theta)$

$\rm \:  =  \: {cosec}^{2}\theta \times {sec}^{2}\theta$

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

$\rm \:  =  \: \dfrac{1}{ {sin}^{2}\theta \: (1 - {sin}^{2}\theta)}$

$\rm \:  =  \: \dfrac{1}{ {sin}^{2}\theta \: - {sin}^{4}\theta}$

Hence, Proved

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Identities Used :-

$\boxed{\tt{ {cosec}^{2}x - {cot}^{2}x = 1 \: }}$

$\boxed{\tt{ {sec}^{2}x - {tan}^{2}x = 1 \: }}$

$\boxed{\tt{ cosecx = \frac{1}{sinx} \: }}$

$\boxed{\tt{ secx = \frac{1}{cosx} \: }}$

$\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1 \: }}$

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More to know :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$