Question

prove that, 1+vosA/1-cosA=(cosecA+cotA)2(square)​

Answer 1

⠀¤ Prove that , $\sf \dfrac{1 + cos A }{ 1 - cos A }\:\:=\:(\: cosec \: A \: + \: cot \: A \:)^2 \:$

$\qquad:\implies \sf \dfrac{1 + cos A }{ 1 - cos A }\:\:=\:(\: cosec \: A \: + \: cot \: A \:)^2 \\\\\\ \qquad:\implies \sf \bigg( \dfrac{ \{\:1 + cos A\:\}\:\{\:1 + cos A\:\}\: }{ \:\{\:1 - cos A\:\}\:\{\:1 + cos A\:\}\: }\:\bigg)\:=\:(\: cosec \: A \: + \: cot \: A \:)^2 \\\\\\ \qquad:\implies \sf \bigg( \dfrac{ \{\:1 + cos A\:\} ^2\:\: }{ \:\{\:1^2 - cos^2 A\:\}\: }\:\bigg)\:=\:(\: cosec \: A \: + \: cot \: A \:)^2 \\\\\\ \qquad:\implies \sf \bigg( \dfrac{ \{\:1 + cos A\:\}^2\:\: }{ \:\{\:1 - cos^2 A\:\}\: }\:\bigg)\:=\:(\: cosec \: A \: + \: cot \: A \:)^2 \\\\\\ \qquad:\implies \sf \bigg( \dfrac{ \{\:1 + cos A\:\}^2\:\: }{ \:\{\:sin^2\:A\:\}\: }\:\bigg)\:=\:(\: cosec \: A \: + \: cot \: A \:)^2 \\\\\\ \qquad:\implies \sf \bigg( \dfrac{ \:1 \:\: }{ \:sin\:A \: }\:+\:\dfrac{ cos \:A }{sin \:A }\bigg)^2\:=\:(\: cosec \: A \: + \: cot \: A \:)^2 \\\\\\ \qquad:\implies \sf \bigg( cosec \:A \:+\:cot \:A \bigg)^2\:=\:(\: cosec \: A \: + \: cot \: A \:)^2 \\\\\\ \qquad:\implies \underline {\pmb{\sf ( cosec \:A \:+\:cot \:A )^2\:=\:(\: cosec \: A \: + \: cot \: A \:)^2}}\:\:\bigstar \\\\\\\qquad\pmb{\purple {\bf Hence\:\:Verified\:!\:}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \underline {\bigstar\pmb{\mathbb{ ADDITIONAL \:\:INFORMATION \::\:}}}\:$

$\dag \:\:\underline { \underline {\purple{\sf Trigonometric \; Indentity \:\:-\: }}}$

$\qquad \sf \: ( I ) \:sin^2\:\theta \:+\:cos^2 \:\:=\:1\:$

$\qquad \sf \: ( II ) \:sin^2\:\theta \:=\: 1 \:-\:cos^2 \:\:\:$

$\qquad \sf \: ( III ) \:\:cos^2 \:\:=\:1 \:-\:sin^2\:\theta \:$

$\qquad \sf \: ( IV ) \:\:\:1 \:+\:cot^2\:\theta \:=\: cosec^2\:\theta \:$

$\qquad \sf \: ( V ) \:\:\:cosec^2\:\theta \:-\:cot^2\:\theta \:=\: 1 \:$

$\qquad \sf \: ( VI ) \:\:\:cosec^2\:\theta \:=\:cot^2\:\theta \:+\: 1 \:$

$\dag \:\:\underline { \underline {\purple{\sf Algebraic \; Indentity \:\:-\: }}}$

$\qquad \sf ( I ) \:\:( a + b)^2 =\:a^2 + b^2 + 2ab \:$

$\qquad \sf ( II ) \:\:( a - b)^2 =\:a^2 + b^2 - 2ab \:$

$\qquad \sf ( III ) \:\: a^2 - b^2 =\:( a + b ) ( a - b ) \:$

$\qquad \sf ( IV ) \:\:( x + b ) ( x + b ) \:=\:x^2 + ( a + b ) x + ab \:$

$\qquad \sf ( V ) \:\: ( a + b + c )^2\:=\: a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \: \:$

$\qquad \sf ( VI ) \:\: ( a + b )^3\:=\: a^3 + b^3 + 3ab ( a + b ) \: \:$