Math, asked by neetuagrawal90, 7 months ago

coseca/coseca-1+coseca/coseca+1=2secsquareA

Answers

Answered by Anonymous
5

✅cosecA/cosecA-1+cosecA/cosecA+1

✅=)cosecA(cosecA-1)+cosec(cosecA+1)

,,,,,,,( cosecA-1)(cosecA+1)

=)cosec^2A-cosecA+cosec^2A+cosecA

✅,,,,,,,,,,,cosec^2A-1

✅=)2cosec^2A/cot^2A

✅=)2(1-cot^2A)/cot^2A

=)2tan^2A-2cot^2A/cot^2A

=)2tan^2A-2

RHS proved....

Answered by SarcasticL0ve
15

\bold{\underline{\underline{\rm{\red{To\;Prove:-}}}}}

\bullet \; \sf{ \dfrac{ \cosec{A}}{ \cosec{A} - 1} + \dfrac{ \cosec{A}}{ \cosec{A} + 1} = 2 \sec^2{A}}

\bold{\underline{\underline{\rm{\pink{Proof:-}}}}}

★ Taking L.H.S. :-

\implies \sf{ \dfrac{ \cosec{A}}{ \cosec{A} - 1} + \dfrac{ \cosec{A}}{ \cosec{A} + 1}}

 \;

\implies \sf{ \dfrac{ \cosec{A}( \cosec{A} - 1) + \cosec{A}( \cosec{A} + 1)}{( \cosec{A} - 1)( \cosec{A} + 1)}}

 \;

\implies \sf{ \dfrac{ \cosec^2{A} - \cancel{ \cosec{A}} + \cosec^2{A} + \cancel{ \cosec{A}}}{( \cosec^2{A} + 1)}}

 \;

\implies \sf{\dfrac{2 \cosec^2{A}}{cot^2{A}}}  \sf{ \bigg(∵ \cosec^2{A} + 1 = \cot^2{A} \bigg)}

 \;

\implies \sf{ \dfrac{ \dfrac{2}{ \cancel{ \sin^2{A}}}}{ \dfrac{ \cos^2{A}}{ \cancel{ \sin^2{A}}}}}

 \;

\implies \sf{ \dfrac{2}{ \cos^2{A}}}

\implies \sf{2 \sec^2{A}}

✦ LHS = RHS

\bold{\underline{\underline{\sf{\purple{\dag \; Hence \; Proved!}}}}}

\rule{200}{2}

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