Question

prove that sinA/1+cosA+1+cosa/sinA=2cosecA

Answer 1

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$\boxed{See \: the \: attachment \: for \: your \: answer.}$

$\large \boxed{ \mathbb{TRIGONOMETRY.}}$

$\huge \bf \underline{ \mathbb{LHS = RHS.}}$

✔✔Hence, it is proved ✅✅.

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Answer 2

here is your answer OK

sinA/1+cosA] +[1+cosA/sinA] = 2cosecA (find LCM for the bolded regions)

=[ (sin2A )+(1+cosA)2 ] / [sinA(1+cosA) ] = 2cosecA

change (1+cosA)2 to a2 +2ab +b2
=[ (sin2A )+(12+2cosA*1 + cos2) ] / [sinA (1+cosA) ] = 2cosecA

change sin2A to 1 - cos2A ( by the 1st identity sin2A + cos2A =1 )
=[ (1 - cos2A )+(1+2cosA + cos2) ] / [sinA (1+cosA) ] = 2cosecA

=1 - cos2A +1+2cosA + cos2 / sinA (1+cosA) = 2cosecA

= 2 +2cosA / sinA (1+cosA) = = 2cosecA

Take 2 OUT
=2 ( 1+ cosA ) / sinA (1+cosA) = 2cosecA

From here ( 1+ cosA ) in numerator & denominator get canceled
=> 2 / sinA = 2cosecA

We know 1 / sinA =cosecA
Thus we get

2cosecA = 2cosecA

hence verified...



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