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Prove 1+secA/secA = sin2A/1-cosA

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Prove 1+secA/secA = sin2A/1-cosA
  • 4 answers

Shreyansh Pandey 3 years, 1 month ago

Very easy

Jaiveer Jat 3 years, 1 month ago

Sin30°

Sanskar Pawar 3 years, 1 month ago

Solve First LHS and then RHS Separately 1+secA / secA = sin2A /1-cosA LHS =1+1/cosA .... sec A can be written as 1/cosA 1+1/cosA /1/cosA Then solve cosA +1 / cosA cos A + 1 /cosA ×cosA/1 Then , cosA and cosA cancel and it remain 1+cosA RHS = sin2A is 1- cosA Then 1- cos2A / cosA 1-cos2A is identity in 9th that is a (a - b ) 2 it is (a+b ) ( a - b ) Then, 1+ cosA × 1- cosA / 1- cos A 1- cosA 1- cos A cancel and it remain 1+ cosA Therefore LHS = RHS Hence , proved

Shubhamnath Keshri 3 years, 1 month ago

Je
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