Question

4. Prove : (sec 0 + tan 0)(1 – sin 0) = cos 0.​

Answer 1

$\Huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

Take L.H.S

$\large {\boxed{\rm{\big( \sec \theta \: + \: \tan \theta \big) \big(1 \: - \: \sin \theta \big) }}} \\ \\ \\ \implies {\sf{\bigg( \dfrac{1}{\cos \theta} \: + \: \dfrac{\sin \theta}{\cos \theta} \bigg) \bigg( 1 \: - \: \sin \theta \bigg)}} \\ \\ \\ \implies {\sf{\bigg( \dfrac{1 \: + \: \sin \theta}{\cos \theta}\bigg) \bigg( 1 \: - \: \sin \theta \bigg)}} \\ \\ \\ \implies {\sf{\bigg( \dfrac{(1 \: + \: \sin \theta)(1 \: - \: \sin \theta}{\cos \theta} \bigg)}} \\ \\ \\ \implies {\sf{\dfrac{1^2 \: - \: \sin ^2 \theta}{\cos \theta}}} \\ \\ \\ \implies {\sf{\dfrac{\cos ^2 \theta}{\cos \theta}}} \\ \\ \\ \implies {\sf{\cos \theta}} \\ \\ \\ \large {\mathbb{HENCE \: PROVED}}$

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$\boxed{\begin{minipage}{6cm}{\large \underline{\textsf{Identities Used}}} \\\\ 1) a^2 - b^2 = (a + b)(a - b) \\\\ 2) \sec \theta = \dfrac{1}{\cos \theta} \\\\ 3) \tan \theta = \dfrac{\sin \theta}{\cos \theta} \end{minipage}}$

Answer 2

Hey!

✔Refer to the given attachment