Question

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Answer 1

$\huge {\mathfrak{ \underline{ \red s \green o \blue l \pink u \purple t \orange i o \red n }}} : -$

$\dfrac{1}{ \tan(A) } - \dfrac{1}{ \tan(2A) }$

$\large \underline {\bold {\boxed{ \tan(2A) = \dfrac{2 \tan(A) }{1 + \tan {}^{2} (A) } }}}$

$\longrightarrow \dfrac{1}{ \tan(A) } - \dfrac{1}{ \dfrac{2 \tan(A) }{1 + \tan {}^{2} (A) } } \\ \\ \longrightarrow \frac{1}{ \tan(A) } - \dfrac{1 + \tan {}^{2} (A) }{2 \tan(A) } \\ \\ \longrightarrow \: \frac{2 - 1 + \tan {}^{2} (A) }{2 \tan(A) } \\ \\ \longrightarrow \: \frac{1 + \tan {}^{2} (A) }{2 \tan(A) }$

$\large {\underline {\bold {\boxed{ \sin(2A) = \frac{2 \tan(A) }{1 + \tan {}^{2} (A) } }}}}$

$\longrightarrow \: \dfrac{1}{ \sin(2A) }$

$\large {\underline {\bold {\boxed{ \cosec(A) = \dfrac{1}{ \sin(A) } }}}}$

$\longrightarrow \: \cosec(2A)$

$\dag \: \huge \bold{ \green{\boxed{ \rm{ \blue{ Hence \: proved}}}}}$

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$\huge \underline \mathfrak{ \red s \blue o \green m \pink e \: \: \red f \green o \blue r \pink m \purple u \orange la \red e } : -$

$(1) \sf \: \: \sin {}^{2} \theta + \cos {}^{2} \theta = 1 \\ \\ (2) \: \cosec {}^{2} \theta - \cot {}^{2} \theta = 1 \\ \\ (3) \: \sec {}^{2} \theta - \tan {}^{2} \theta = 1 \\ \\ (4) \cos2 \theta = 2 \cos {}^{2} \theta - 1 \\ \\ (5) \sin 2\theta = 2 \sin\theta\cos\theta$

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Answer 2

Step-by-step explanation:

Given :

Area of rhombus = 120 cm²

Length of diagonal = 8 cm

To find :

Length of another diagonal

According to the question,

$\sf{ : \implies Area \: of \: rhombus = \dfrac{1}{2} \times d _{1} \times d_{2} }$

$\sf : \implies{ {120 \: cm}^{2} = \dfrac{1}{2} \times 8 \: cm \times x}$

$\sf : \implies{ {120 \: cm}^{2} \times 2 = 8 \: cm \times x }$

$\sf : \implies{ {240 \: cm}^{2} = 8 \: cm \times x}$

$\sf : \implies{ \dfrac{240}{8} \: cm = x}$

${ \underline{ \boxed{ \sf \pink{ : \implies{ \bm3 \bm0 \: c m =x}}}}}$

${ \therefore{ \underline{\sf{So \:,the \: length \: of \: other \: diagonal \: is \: 3 0 \: cm}}}}$