Question

If m sinA= n cosA , prove tanA + cotA/tanA -cotA =n sinA + m cosA/n sinA - m cos A = n2 +m2/n2 - m2

Answer 1

Solution -

Given: 

msin A = ncosA

Therefore, sinA/cosA = n/m = tanA  and  cosA/sinA = m/n = cotA

Now,

LHS = (tanA + cotA) / (tanA - cotA) 

Substitute the value of tanA ans cotA

LHS = (n/m + m/n) / (n/m - m/n)
 
       = ((n^2 + m^2) / mn) / ((n^2 - m^2) / mn)
 
       = (n^2 + m^2) / (n^2 - m^2)
    
       = RHS

Also,

LHS = (nsinA + mcosA / nsinA - mcosA)

Divide numerator and denominator by cosA

LHS = (ntanA + m) / (ntanA - m)

Substitute the value of tanA = n/m

LHS = (n * n/m + m) / n * n/m - m)
   
       = ((n^2 + m^2) / m) / ((n^2 - m^2) / m)
 
        = (n^2 + m^2) / (n^2 - m^2)
 
        = RHS

Hence proved.... tanA + cotA/tanA -cotA =n sinA + m cosA/n sinA - m cos A = (n^2 +m^2) / (n^2 - m^2)

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