Question

$\displaystyle \: \sf( \cosec x - \sin \: x)( \sec x - \cos x) = \frac{1}{ \tan \: x + \cot x}$
$\sf \: No \: Spam$​

Answer 1

“Refer the attachment”

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Answer 2

$\boxed{ \boxed{\begin{array}{cc} \rightarrow\sf Given \\ \hookrightarrow\rm( \cosec x - \sin \: x)( \sec x - \cos x) = \dfrac{1}{ \tan \: x + \cot x} \end{array}}}$

$\boxed{ \boxed{\begin{array}{cc} \rightarrow\sf To \: Prove\\ \hookrightarrow\rm( \cosec x - \sin \: x)( \sec x - \cos x) = \dfrac{1}{ \tan \: x + \cot x} = 1 \end{array}}}$

$\boxed{ \boxed{\begin{array}{ll} \hookrightarrow\rm( \cosec x - \sin \: x)( \sec x - \cos x) = \dfrac{1}{ \tan \: x + \cot x} \\ \\ \\ \hookrightarrow \bigg\lbrack \rm \dfrac{1}{ \sin x} - \sin x \bigg \rbrack \bigg \lbrack \dfrac{1}{ \cos x} - \cos x \bigg \rbrack \left\lbrack \dfrac{1}{ \dfrac{ \sin x}{ \cos x} + \dfrac{ \cos x}{ \sin x} }\right \rbrack \\ \\ \\ \hookrightarrow \bigg\lbrack \rm \dfrac{1- \sin^{2} x}{ \sin x} \bigg \rbrack \bigg \lbrack \dfrac{1- \cos^{2} x}{ \cos x} \bigg \rbrack \left\lbrack { \dfrac{ \cos x}{ \sin x} + \dfrac{ \sin x}{ \cos x} }\right \rbrack \\ \\ \\ \hookrightarrow \bigg\lbrack \rm \dfrac{\cos^{2} x}{ \sin x} \bigg \rbrack \bigg \lbrack \dfrac{ \sin^{2} x}{ \cos x} \bigg \rbrack \left\lbrack { \dfrac{ \cos ^{2} x + \sin^{2}x }{ \sin x . \cos x} }\right \rbrack \\ \\ \\ \hookrightarrow \rm\dfrac{ \cancel{\cos ^{2} x}}{ \cancel{\sin x}} \times \dfrac{ \cancel{ \sin ^{2} x}}{ \cancel{\cos x}} \times \dfrac{ 1}{ \cos x. \sin x} \\ \\ \\ \hookrightarrow \rm \dfrac{ \cos x. \sin x}{1} \times \dfrac{1}{ \cos x. \sin x} \\ \\ \\ \hookrightarrow \rm\cancel\dfrac{\cos x. \sin x}{\cos x. \sin x} \\ \\ \\ \hookrightarrow 1\end{array}}}$