How I can prove 1/(cosecA-cot)=cosecA+cotA. ?
Answers
Step-by-step explanation:
LHS
(1/cosecA-cotA)-(1/sinA)
={1/(1/sinA-cosA/sinA)}-(1/sinA)
=[1/{(1-cosA)/sinA}]-(1/sinA)
=sinA/(1-cosA)-(1/sinA)
=(sin²A-1+cosA)/sinA(1-cosA)
={(1-cos²A)-(1-cosA)}/sinA(1-cosA)
={(1+cosA)(1-cosA)-(1-cosA)}/sinA(1-cosA)
=(1-cosA)(1+cosA-1)/sinA(1-cosA)
=cosA/sinA
=cotA
RHS
(1/sinA)-(1/cosecA+cotA)
=(1/sinA)-{1/(1/sinA+cosA/sinA)}
=(1/sinA)-1/{(1+cosA)/sinA}
=(1/sinA)-sinA/(1+cosA)
=(1+cosA-sin²A)/sinA(1+cosA)
={1+cosA-(1-cos²A)}/{sinA(1+cosA)}
={(1+cosA)-(1+cosA)(1-cosA)}/{sinA(1+cosA)}
=(1+cosA)(1-1+cosA)/sinA(1+cosA)
=cosA/sinA
=cotA
∴, LHS=RHS
Answer:
Step-by-step explanation:.
LHS= 1/(cosecA-cotA)
Now cosecA can be written as 1/sinA and cotA can be written as cosA/sinA
LHS=1/{(1/sinA)-(cosA/sinA)}
=sinA/(1-cosA)
Now multiplying and dividing by sinA
=1/(cosecA-cotA)