Question

1-cosA/1+cosA=(cosecA-cotA)2​

Answer 1

To Prove :-

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

Solution :-

$\sf L.H.S = \dfrac{1-cosA}{1+cosA}$

Multiply the numerator and denominator by $\sf (1-cosA)$.

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

        $\sf = \dfrac{(1-cosA)^2}{1-cos^2A}$

$\bf\dag \ 1- cos^2A = sin^2A$

       $\sf = \dfrac{(1-cosA)^2}{sin^2A}$

       $= \sf \left( \dfrac{1-cosA}{sinA} \right)^2$

       $= \sf \left( \dfrac{1}{sinA} - \dfrac{cosA}{sinA} \right)^2$

$\bf\dag \ cosecA = \dfrac{1}{sinA}$

$\bf\dag \ cotA = \dfrac{cosA}{sinA}$

       $= \sf (cosecA-cotA)^2$

       $= \sf R.H.S$

Hence proved.

       

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: Answer:-}}}}}$

_______________________________________________★

◉TO PROVE:-

$\tt \bold{ \frac{1- cosA}{1 + cosA } = {(cosec A \: - cotA)}^{2} }$

◉ FORMULA USED:-

• $Cosec A = \frac{1}{sinA}$
• $Cot A = \frac{Cos A }{ SinA }$
• $sin^2\theta+cos^2\theta=1$

${\boxed{\underline{\bf{\red{R .H.S}}}}}$

$\leadsto \: {(cosec A \: - cotA)}^{2}$

$\leadsto \: {(\frac{1}{sinA} - \frac{Cos A }{ SinA })}^{2}$

$\leadsto \: {( \frac{1 - \: Cos A }{ SinA\: })}^{2}$

$\leadsto \: \frac{ {(1 - \: Cos A)}^{2} }{ \: \: { \:Sin }^{2} A}$

$\leadsto \frac{ {(1 - cos A)}^{2} }{1 - {cos}^{2} A }$

$\leadsto \: \frac{( \cancel{1 - cos A})(1 - cos A)}{( \cancel{1 - cos A})(1 + cos A)}$

$\leadsto \bf \pink{ \frac{(1 - cos A)}{(1 + cos A)} }$

${\boxed{\underline{\bf{\red{ L.H.S}}}}}$

THEREFORE, L.H.S = R.H.S

HOPE IT HELPS