If $ A+B=90^{\circ}, $ then prove that $ \sqrt{\frac{t a n A \text

1.	If $ A+B=90^{\circ}, $ then prove that $ \sqrt{\frac{t a n A \text { tan } B+t a n A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} B}}=\tan A $
Answer» LHS = root tan a tan b + tan a cot b/sin a sec b - sin^2 b/cos^2 a = root tan a tan(90-a) + tan a cot(90-a)/sin a sec(90-a) - sin^2(90-a)/cos^2 a = root tan a.cot a + tan a.tana/sin a.cosec a - cos^2 a/cos^2 a = root 1+tan^2 a/1 - 1 = root tan^2 a = tan a.

If $ A+B=90^{\circ}, $ then prove that $ \sqrt{\frac{t a n A \text { tan } B+t a n A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} B}}=\tan A $

Answer»

LHS = root tan a tan b + tan a cot b/sin a sec b - sin^2 b/cos^2 a

= root tan a tan(90-a) + tan a cot(90-a)/sin a sec(90-a) - sin^2(90-a)/cos^2 a

= root tan a.cot a + tan a.tana/sin a.cosec a - cos^2 a/cos^2 a

= root 1+tan^2 a/1 - 1

= root tan^2 a

= tan a.