Question

(1-tanA÷1-cotA)^2 = tan^2A

Answer 1

(1-tanA)square / (1-cotA) square
(1-tanA) square / [(tanA - 1)/ tanA)] square
(1-tanA) square / [(tanA - 1)sq / tan sq A]
(1-tanA)sq / [(1-tanA) sq / tan sq A]
Tan sq A

Answer 2

Solution:

$LHS = \left(\frac{1-tanA}{1-cotA}\right)^{2}$

=$\left(\frac{1-tanA}{1-\frac{1}{tanA}}\right)^{2}$ /* cotA =1/tanA */

=$\left(\frac{1-tanA}{\frac{(tanA-1)}{tanA}}\right)^{2}$

=$\left(\frac{tanA(1-tanA)}{-(1-tanA)}\right)^{2}$

/* After cancellation , we get

=$(-tanA)^{2}$

=$tan^{2}A$

= $RHS$

Therefore,

$\left(\frac{1-tanA}{1-cotA}\right)^{2}$=$tan^{2}A$

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