Question

tanA          cotA
---------- + -------------- =  1+cosecA×secA                                                                             
1-cotA      1-tan A
PROVE

Answer 1

Formula:
(1) $( x^{3} - y^{3}) = (x - y) ( x^{2} + (x)(y) + y^{2})$
(2) $sin^{2}(\theta) + cos^{2}(\theta) = 1$
(3) $tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$
(4) $cot(\theta) = \frac{cos(\theta)}{sin(\theta)}$
(5) $sec(\theta) = \frac{1}{cos(\theta)}$
(6) $cosec(\theta) = \frac{1}{sin(\theta)}$


$\frac{tan(A)}{1 - cot(A)} = \frac{ \frac{sin(A)}{cos(A)} }{1 - \frac{cos(A)}{sin(A)} } = \frac{sin^{2}(A)}{cos(A)(sin(A) - cos(A))}$

$\frac{cot(A)}{1 - tan(A)} = \frac{ \frac{cos(A)}{sin(A)} }{1 - \frac{sin(A)}{cos(A)} } = \frac{cos^{2}(A)}{sin(A)(cos(A) - sin(A))}$

Thus,
$\frac{tan(A)}{1 - cot(A)} +\frac{cot(A)}{1 - tan(A)}\\ \\ =\frac{sin^{2}(A)}{cos(A)(sin(A) - cos(A))} +\frac{cos^{2}(A)}{sin(A)(cos(A) - sin(A))}\\ \\ =\frac{sin^{2}(A)}{cos(A)(sin(A) - cos(A))} - \frac{cos^{2}(A)}{sin(A)(sin(A) - cos(A))}\\ \\ =\frac{sin^{3}(A)}{sin(A)cos(A)(sin(A) - cos(A))} - \frac{cos^{3}(A)}{sin(A)cos(A)(sin(A) - cos(A))}$$\\ \\ =\frac{sin^{3}(A) - cos^{3}(A)}{sin(A)cos(A)(sin(A) - cos(A))}\\ \\ =\frac{((sin(A) - cos(A)))(sin^{2}(A) + sin(A)cos(A)+cos^{2}(A))}{sin(A)cos(A)(sin(A) - cos(A))}\\ \\ =\frac{((sin(A) - cos(A)))(1 + sin(A)cos(A))}{sin(A)cos(A)(sin(A) - cos(A))}\\ \\ =\frac{(1 + sin(A)cos(A))}{sin(A)cos(A)} \\ \\ =1 + \frac{1}{sin(A)cos(A)}\\ \\ =1 + cosec(A)sec(A)$

Answer 2

[tex] \frac{tan\ A}{1-\frac{1}{tan\ A}} + \frac{1/tan\ A}{1 - tan\ A} \\ \\ \frac{tan^{2}\ A}{tan\ A -1} + \frac{1}{tan\ A\ (1 - tan\ A)} \\ \\ \frac{- tan^{3}\ A +1}{tan\ A\ (1 - tan\ A)} \\ \\ \frac{(1-tan\ A)(1+tan\ A + tan^{2}\ A)}{tan\ A (1-tan\ A)} \\ \\ 1 + cot\ A + tan\ A \\ \\ 1 + \frac{cos^{2}\ A + sin^{2}\ A}{cos\ A\ sin\ A} \\ \\ Hence\ the\ answer\ on\ the\ RHS \\ [/tex]