Question

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Answer 1

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies \frac{1 + sec \: \theta}{sec \: \theta} = \frac{ {sin}^{2} \: \theta}{1 - cos \: \theta} \\ \\ \red{\underline \bold{To \: Prove:}} \\ \tt: \implies \frac{1 + sec \: \theta}{sec \: \theta} = \frac{ {sin}^{2} \: \theta}{1 - cos \: \theta}$

• According to given question :

$\tt: \implies \frac{1 + sec \: \theta}{sec \: \theta} = \frac{ {sin}^{2} \: \theta}{1 - cos \: \theta} \\ \\ \bold{As \: we \: know \: that} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{L.H.S} \\ \tt: \implies \frac{1 + sec \: \theta}{sec \: \theta} \\\\ \tt\circ\:sec\:\theta=\frac{1}{cos\:\theta}\\ \\ \tt: \implies \frac{1 + \frac{1}{cos \: \theta} }{ \frac{1}{cos \: \theta} } \\ \\ \tt: \implies \frac{ \frac{cos \: \theta + 1}{cos \: \theta} }{ \frac{1}{cos \: \theta} } \\ \\ \tt: \implies \frac{cos^{2} + cos \: \theta}{cos \: \theta} \\ \\ \tt: \implies cos \: \theta + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \bold{R.H.S} \\ \tt: \implies \frac{ {sin}^{2} \: \theta }{1 - cos \: \theta} \\\\ \tt\circ\:sin^{2}\:\theta+cos^{2}\:\theta=1 \\\\ \tt: \implies \frac{1 - {cos}^{2} \: \theta}{1 - cos \: \theta} \\ \\ \tt: \implies \frac{ {1}^{2} - {cos}^{2} \: \theta}{1 - cos \: \theta} \\\\ \tt\circ\:a^{2}-b^{2}=(a+b)(a-b) \\\\ \tt: \implies \frac{(1 + cos \: \theta)(1 - cos \: \theta)}{1 - cos \: \theta} \\ \\ \tt: \implies cos \: \theta + 1 \\ \\ \green{ \huge{\tt\therefore L.H.S =R.H.S}} \\ \\ \: \: \: \: \: \: \blue{ \huge{\bold{Proved }}}$

Answer 2

PROOF :

$</p><p>\begin{lgathered}\tt: \implies \frac{1 + sec \: \theta}{sec \: \theta} = \frac{ {sin}^{2} \: \theta}{1 - cos \: \theta} \\ \\ \bold{As \: we \: know \: that} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{L.H.S} \\ \tt: \implies \frac{1 + sec \: \theta}{sec \: \theta} \\\\ \tt\circ\:sec\:\theta=\frac{1}{cos\:\theta}\\ \\ \tt: \implies \frac{1 + \frac{1}{cos \: \theta} }{ \frac{1}{cos \: \theta} } \\ \\ \tt: \implies \frac{ \frac{cos \: \theta + 1}{cos \: \theta} }{ \frac{1}{cos \: \theta} } \\ \\ \tt: \implies \frac{cos^{2} + cos \: \theta}{cos \: \theta} \\ \\ \tt: \implies cos \: \theta + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \bold{R.H.S} \\ \tt: \implies \frac{ {sin}^{2} \: \theta }{1 - cos \: \theta} \\\\ \tt\circ\:sin^{2}\:\theta+cos^{2}\:\theta=1 \\\\ \tt: \implies \frac{1 - {cos}^{2} \: \theta}{1 - cos \: \theta} \\ \\ \tt: \implies \frac{ {1}^{2} - {cos}^{2} \: \theta}{1 - cos \: \theta} \\\\ \tt\circ\:a^{2}-b^{2}=(a+b)(a-b) \\\\ \tt: \implies \frac{(1 + cos \: \theta)(1 - cos \: \theta)}{1 - cos \: \theta} \\ \\ \tt: \implies cos \: \theta + 1 \\ \\ \orange{ \huge{\tt\therefore L.H.S =R.H.S}} \\ \\ \: \: \: \: \: \: \purple{ \huge{\bold{Proved }}}\end{lgathered}$