1+cos theta÷ sin theta+ sin theta ÷ 1+cos theta
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we have to prove that (1+cosA)/sinA + sinA/(1+cosA) = 2 cosec A
we have ,
LHS = (1+cosA)/sinA + sinA/(1+cosA)
=>[ (1+cosA)²+sin²A]/sinA(1+cosA) [taking LCM]
=>[ 1+cos²A+2cosA+sin²A]/sinA(1+cosA)
=> [1+1+2cosA]/sinA (1+cosA) [since sin²A+cos²A=1]
=> [2+2cosA]/sinA (1+cosA)
=> 2(1+cosA)/sinA (1+cosA) [taken 2 as common]
=>2/sinA =>2cosecA
LHS=RHS
hope this helped you
we have ,
LHS = (1+cosA)/sinA + sinA/(1+cosA)
=>[ (1+cosA)²+sin²A]/sinA(1+cosA) [taking LCM]
=>[ 1+cos²A+2cosA+sin²A]/sinA(1+cosA)
=> [1+1+2cosA]/sinA (1+cosA) [since sin²A+cos²A=1]
=> [2+2cosA]/sinA (1+cosA)
=> 2(1+cosA)/sinA (1+cosA) [taken 2 as common]
=>2/sinA =>2cosecA
LHS=RHS
hope this helped you
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