Question

If tan α - tan β = m, cot α - cot β = n, then prove that cot(α - β) = $\frac{1}{m} - \frac{1}{n}$

Answer 1

we have to prove : $cot(\alpha-\beta)=\frac{1}{m}-\frac{1}{n}$

given, $cot\alpha-cot\beta=n$

$\frac{1}{tan\alpha}-\frac{1}{tan\beta}=n$

$\frac{tan\beta-tan\alpha}{tan\alpha tan\beta}=n$

$\frac{-(tan\alpha-tan\beta)}{tan\alpha tan\beta}=n$

$\frac{-m}{n}=tan\alpha tan\beta$....(1)

now, LHS = $cos(\alpha-\beta)$

= $\frac{1}{tan(\alpha-\beta)}$

= $\frac{1+tan\alpha.tan\beta}{tan\alpha-tan\beta}$

= { 1 + (-m/n)}/(m)

= 1/m + (-m/n)/m

= 1/m - 1/n = RHS

Answer 2

HELLO DEAR,


GIVEN:-
tanα - tanβ = m---------( 1 )

cotα - cotβ = n

=> 1/tanα - 1/tanβ = n

=> (tanβ - tanα)/(tanα.tanβ) = n

=> -m/(tanα.tanβ) = n [From----( 1 ) ]

=> -m/n = tanα.tanβ----------( 2 )


now, cot(α - β) = 1/tan(α - β)

=> 1/[(tanα - tanβ)/(1 + tanα.tanβ)]

=> 1/[ {m}/{1 + (-m/n)}]

=> 1/[mn/(n - m)]

=> (n - m)/(mn)

=> 1/m - 1/n


I HOPE IT'S HELP YOU DEAR,
THANKS