Question

if tan A-tan B=m and cot A-cot B =n then prove that cot(A-B) =1/m-1/n​

Answer 1

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cot A - cotB = n

=>  1/ tan A  - 1/tan B = n

=> (tanB - tan A )/ tanA tanB = n

=> -m/ tanA tanB = n

=> tanA tanB = -m/n 

now,

cot (B - A) = (cotA cotB + 1)/ cotB - cotA

= 1 + tanA tanB    x       1                 

   tanA tanB              cotB - cot A

=  1+ (-m/ n)    x    1 

    (-m/n)              (-n)

= n-m     x 1

    m          n

=n-m

 nm 

= 1 /m  - 1/n 

HENCE PROVED