Question

If A+ B = 90°, then prove that $\sqrt{ \frac{tan A .tan B - tan A. cot B}{sin A. sec B} - \frac{sin^2 B}{cos^2 A} }$ = tan A

Answer 1

Some relations you should know:
If A+B=90°, then:
→ tanB = cotA
→ tanA = cotB 
→ sinB = cosA
→ sinA = cosB 

$\sqrt{ \frac{tanA.tanB-tanA.cotB}{sinA.secB}- \frac{sin^2B}{cos^2A} } \\ \\ = \sqrt{ \frac{tanA.cotA-tanA.tanA}{sinA.(1/cosB)}- \frac{sin^2B}{sin^2B} } \\ \\ =\sqrt{ \frac{tanA.(1/tanA)-tan^2A}{sinA.(1/sinA)}- \frac{sin^2B}{sin^2B} } \\ \\ =\sqrt{ \frac{1-tan^2A}{1}-1 }\\ \\ = \sqrt{-tan^2A} \\ \\=\pm i\ tanA$

It's not coming tanA. 

Answer 2

We have ,  a + b = 90°
 ⇒ b = 90° - a  ................(1)