Question

prove that sin(90-0)/cosec(90-0)-cot (90-0) = cot (90-0)​

Answer 1

Given :     Sin (90 - θ )/ Cosec(90 - θ)   - Cot(90 - θ)  = Cot(90 - θ)  

To find :  Prove  Sin (90 - θ )/ Cosec(90 - θ)  -  Cot(90 - θ) = Cot(90 - θ)  

Solution:

There seems mistake  in Data  

lets put

θ  = 45°

LHS

= Sin45 / Cosec45  - Cot45

= 1/2 -  1

= -1/2

RHS =  Cot45 =  1  

LHS  ≠ RHS

There is mistake in Data

Even if we take

Sin (90 - θ )/ (Cosec(90 - θ)   - Cot(90 - θ) ) = Cot(90 - θ)  

θ  = 45°

LHS = (1/√2)/ (√2  - 1)   =    (√2  + 1)  / √2  = 1  +  1/√2

RHS = 1

LHS ≠ RHS

Hence Mistake in Data

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