Question

plzz do it..........​

Answer 1

✧ $\sf \dfrac{1 + cosA}{1 - cosA} = (cosecA + cotA)^{2}$

LHS:-

$\sf \dfrac{1 + cosA}{1 - cosA}$

By rationalisation:-

$\sf \dfrac{1 + cosA}{1 - cosA} \times \dfrac{1 + cosA}{1 + cosA}$

$\sf ➝ \: \dfrac{(1 + cosA)(1 + cosA)}{(1 - cosA)(1 + cosA)}$

$\sf \color{orange}By \: using \: algebraic \: identities:-$

$\sf \color{pink} (a + b) (a + b) = (a + b)²$

$\sf \color{fuchsia} (a + b) (a - b) = a² - b²$

$\sf ➝ \: \dfrac{(1 + cosA)^{2}}{{(1)}^{2} - {(cosA)}^{2}}$

$\sf ➝ \: \dfrac{(1 + cosA)^{2}}{1 - cos²A}$

$\sf \color{fuchsia} Now, \: we \: know \: that \: 1 - cos²A = sin²A$

$\sf ➝ \: \dfrac{(1 + cosA)^2}{sin²A}$

$\sf ➝ \: (\dfrac{1 + cosA}{sinA})^{2}$

$\sf ➝ \: (\dfrac{1}{sinA} + \dfrac{cosA}{sinA})^{2}$

$\sf \color{green} As, \: We \: know \: that,$

$\sf \color{fuchsia} \dfrac{1}{sinA} = cosecA$

$\sf \color{red} \dfrac{cosA}{sinA} = cotA$

$\sf ➝ \: (\dfrac{1}{sinA} + \dfrac{cosA}{sinA})^{2} = (cosecA + cotA)^{2}$

RHS:-

$\sf (cosecA + cotA)^{2}$

$\sf \therefore LHS = RHS$

$\sf Hence, \: Proved$

Answer 2

$\mathsf\red{ \frac{1 + \cos \:A }{1 - \cos \: A} = {( \cosec A+ \cot \: A)}^{2} } \\ \\ \mathsf{RHS \: = \: {( \cosec A+ \cot \: A)}^{2}} \\ \\ \mathsf{ = { \Big( \frac{1}{ \sin \:A } + \frac{ \cos \: A}{ \sin \:A } \Big) }^{2} } \\ \\ \mathsf{ = \: { \Big(\frac{1 + \cos \: A }{sin \: A} }^{2} \Big) \: \: \: \: \: \: \: \: } \\ \\ \mathsf{ = \frac{ {(1 + cos \: A)}^{2} }{ {sin}^{2}A } \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \mathsf{ = \: \frac{ {(1 + cos \:A )}^{2} }{ {1 - {cos}^{2} A} \: } \: \: \: \: \: \: \: \: \: \: } \\ \\ \mathsf{ = \: \frac{ {(1 + cos \:A).(1 + cos \: A)} }{(1 + cos \: A).(1 - cosA)} } \\ \\ \mathsf{ = \frac{1 + cos \:A }{1 - cos \: A} } \\ \\ \mathsf{ \: = \: LHS} \\ \\ \boxed{\mathsf \red{ \therefore \:LHS \: = \: RHS}}$

$\large\mathsf{Hence \: proved}$

$\mathsf{}$

${\tt{\fcolorbox{black}{pink}{© \: ITZBFF}}}$