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Prove that 1/(cosec0 - cot0) - 1/(sin0) = 1/(sin0) - 1/(cosec0 + cot0)

Prove that 1/(cosec0 - cot0) - 1/(sin0) =  1/(sin0) - 1/(cosec0 + cot0)

Grade:10

2 Answers

Arun
25750 Points
6 years ago
LHS=1/(cosecA-cotA) -1/sinA
=(cosecA+cotA)/cosec^2A-cot^2A) -cosecA
=cosecA+cotA-cosecA
=cotA
RHS=1/sinA-1/(cosecA+cotA)
=cosecA-(cosecA-cotA)/(cosec^2A-cot^2A)
=cosecA-(cosecA-cotA)
=cotA
LHS=RHS
Regards
Arun (askIITians forum expert)
Vedant
35 Points
6 years ago
First let us simplify the equation and see where it goes
1/(cosec0 - cot0) - 1/(sin0) =  1/(sin0) - 1/(cosec0 + cot0)
1/(cosec0 + cot0)On transposing we find that,2/(sin0) = 1/(cosec0 - cot0) +
(cosec0 - cot0)[(cosec0 - cot0)+(cosec0 - cot0)]/(cosec0 + cot0)2/(sin0) =
(cosec20 - cot20) (a + b)(a – b) = (a2 – b2)(cosec0)/22/(sin0) =
(cosec20 - cot20)(sin0)(cosec0) =

As we know that sin0 and cosec0 are reciprocals of each other, they will cancel out to give 1
Therefore,
(cosec20 - cot20) = 1
1/(sin20) – (cos20)/(sin20) = 1
1 – (cos20)/(sin20) = 1
Therefore,
1 – (cos20) must be equal to sin20
or
(cos20) must be equal to 1 for this equation to be true(sin20) +
As we all know that this is a famous trignometric identity, this equation must be true
 
 
 
 
 
HENCE PROVED, 1/(cosec0 - cot0) - 1/(sin0) =  1/(sin0) - 1/(cosec0 + cot0)

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