Question

Prove that LHS =RHS
plz qnswer fast with proper steps and explanation....​

Answer 1

Answer:

ans is in ans ..hope it will help

Answer 2

$\boxed{ \red{\sf \: Let's \: Prove \: \: \sqrt{ \dfrac{1 + \sin{A }}{1 - \sin{A}} } \: = \: \sec{A} \: + \: \tan{A}}}$

$\bf \: \purple{Let \: us \: firstly \: solve \: (LHS) \: Left \: hand \: side. }$

$\sqrt{ \dfrac{1 + \sin{A }}{1 - \sin{A}} }$

$\sf \: \dfrac{ \sqrt{1 + \sin{A }} }{ \sqrt{ 1 - \sin{A}}} \: \times \: \dfrac{ \sqrt{1 + \sin{A }} }{ \sqrt{1 + \sin{A}} }$

$\boxed{\sf \red{(a - b) \: (a+b) \: = \: a ^{2} - b ^{2} }}$

$\sf \: \dfrac{ \sqrt{(1 + \sin{A}) ^{2} } }{ \sqrt{ (1)^{2} - ( \sin {^2}A) }} \:$

$\sf \: By \: cancelling \: root \: with \: square \: in \: numerater.$

$\sf \: \dfrac{1 + \sin{A} }{ \sqrt{(1) ^{2} - \sin^{2}A } }$

$\sf \: We \: know \: that , \: (1)^{2} \: = \: 1$

$\sf \: \dfrac{1 + \sin{} }{ \sqrt{1 - \sin^{2} A } }$

$\sf \: \dfrac{1 + \sin{} }{ \sqrt{ \cos^{2} A } }$

$\sf \: By \: cancelling \: root \: with \: square \: in \: numerater.$

$\sf \: \dfrac{1 + \sin{A} }{ \cos{A} }$

$\bf \: \purple{ Now, \: We \: will \: solve \: (RHS) \: Right \: Hand \: side : }$

$\sf \: \sec{A} \: + \: \tan{A}$

$\sf \dfrac{1}{ \cos{A}} \: + \: \dfrac{ \sin{A} }{\cos{A}}$

$\sf \: \dfrac{1 + \sin{A} }{ \cos{A} }$

$\begin{gathered} \\ \boxed{\bf\:L.H.S.= \:R.H.S.}\\\end{gathered}$

$\begin{gathered} \\ \bf{ \green{\underbrace{\red{Hence \: \: Proved}}}}\\\end{gathered}$

$\begin{gathered}\dag\: \green{\textsf{\textbf{ \purple{More to know : }}}} \\ \end{gathered}$

$\sf \sin^{2} {A} \: + \: \cos^{2} {A} \: = \: 1$

$\sf \: 1 - \cos {^{2} {A}} \: = \: \sin{^{2} {A}}$

$\sf \: 1 - \sin{^{2} {A}} \: = \: \ \cos {^{2} {A}}$