Question

if A+B=90° , then (tanA*tanB+tanA*cotB/sinA*secB)-(sin²B/cos²A)​

Answer 1

Answer:

here is yr answer

Step-by-step explanation:

I hope this helps u

Answer 2

$<p style="color:violet;font-family:cursive;background :black;font-size:25px;">Given, LHS = root tan a tan b + tan a cot b/sin a sec b - sin^2 b/cos^2 a</p><p></p><p>      = root tan a tan(90-a) + tan a cot(90-a)/sin a sec(90-a) - sin^2(90-a)/cos^2 a</p><p></p><p>      = root tan a.cot a + tan a.tana/sin a.cosec a - cos^2 a/cos^2 a</p><p></p><p>      = root 1+tan^2 a/1 - 1</p><p></p><p>     = root tan^2 a</p><p></p><p>    = tan a.</p><p></p><p></p><p>.</p>$