Question

Trignometry based question pls solve as soon as possible.. Thanks!!

Answer 1

hello,

LHS=
[(sinA/1+cosA)+(1+cosA)/sinA]·[(sinA/1-cosA)-(1-cosA)/sinA]
taking LCM,

⇒[sin²A+(1+cosA)²/(1+cosA)(sinA)]·[sin²A-(1-cosA)²/(1-cosA)(sinA)]

⇒[(sin²A+1+cos²A+2cosA)/(1+cosA)(sinA)·[sin²A-(1+cos²A-2cosA)/(1-cosA)(sinA)]    opening bracket using whole square identity

⇒[(2+2cosA)/(1+cosA)(sinA)]·[(sin²A-1-cos²A+2cosA)/(1-cosA)(sinA)]
                        sin²A+cos²A=1

⇒[2(1+cosA)/(1+cosA)(sinA)]·[(-cos²A-cos²A+2cosA)/(1-cosA)(sinA)]

⇒[2/sinA]·[(-2cos²A+2cosA)/(1-cosA)(sinA)]

⇒2cosecA·[2cosA(1-cosA)/(1-cosA)(sinA)]

⇒2cosecA·[2cosA/sinA]

⇒2cosecA·2cotA         cosA/sinA=cotA

⇒4.cosecA.cotA
=RHS
hence proved
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