Question

it's urgent guys plzz answer this.....

Answer 1

Here is the answer, I got A and B as 45° each!!

Answer 2

$Given, A+B=90\\\\LHS=\sqrt{\frac{tanA.tanB+tanB.tanA}{sinA.secB}-\frac{sin^2B}{cos^2A}}\\\\\\A+B=90\\tan(90-A)=tanB\\cotA=tanB\\and, sinB=cosA\\\\=\sqrt{\frac{cotB.tanB+tanB.cotB}{cosB.secB}-\frac{cos^2A}{cos^2A}}\\\sqrt{2-1}=1 \neq TanA=RHS\\\\hence, LHS \neq RHS$