May 2, 2018

Given that $A + B = {90}^{\circ}$ then $A = 90 - {B}^{\circ}$

$\rightarrow \frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{{\sin}^{2} B}{{\cos}^{2} A}$

=(tanA[tanB+cotB])/(sinAsecB)-(sin^2B)/(cos^2(90^@-B)

$= \frac{\left(\frac{\cancel{\sin A}}{\cos} A\right) \left[\sin \frac{B}{\cos} B + \cos \frac{B}{\sin} B\right]}{\frac{\cancel{\sin A}}{\cos} B} - \frac{{\sin}^{2} B}{{\sin}^{2} B}$

$= \frac{\left(\frac{1}{\cos} A\right) \left[\frac{{\sin}^{2} B + {\cos}^{2} B}{\sin B \cdot \cancel{\cos B}}\right]}{\frac{1}{\cancel{\cos B}}} - 1$

$= \frac{1}{\cos \left({90}^{\circ} - B\right) \sin B} - 1$

$= \frac{1}{\sin} ^ 2 B - 1 = \frac{1 - {\sin}^{2} B}{\sin} ^ 2 B = \frac{{\cos}^{2} B}{{\sin}^{2} B} = {\cot}^{2} B$

May 3, 2018

Given that $A + B = {90}^{\circ}$ then A=90-B^@#
Now

$\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{{\sin}^{2} B}{{\cos}^{2} A}$

$\frac{\tan \left({90}^{\circ} - B\right) \tan B + \tan \left({90}^{\circ} - B\right) \cot B}{\sin \left({90}^{\circ} - B\right) \sec B} - \frac{{\sin}^{2} B}{{\cos}^{2} \left({90}^{\circ} - B\right)}$

$= \frac{\cot B \tan B + \cot B \cot B}{\cos B \sec B} - \frac{{\sin}^{2} B}{{\sin}^{2} B}$

$= 1 + {\cot}^{2} B - 1 = {\cot}^{2} B$