Question

plz help me out........guys it will give u 100 points.

Answer 1

Step-by-step explanation:

Given, A + B = 90° (or) B = 90° - A.

LHS:

$=\sqrt{\frac{tanAtanB+tanAcotB}{sinAsecB} - \frac{sin^2B}{cos^2A}}$

$=\sqrt{\frac{tanAtan(90-A)+tanAcot(90-A)}{sinAsec(90-A)}-\frac{sin^2(90-A)}{cos^2A}}$

$=\sqrt{\frac{tanAcotA+tanAtanA}{sinAcosecA}-\frac{cos^2A}{cos^2A}}$

$=\sqrt{\frac{tanA*\frac{1}{tanA}+tan^2A}{sinA*\frac{1}{sinA}}-\frac{cos^2A}{cos^2A}}$

$=\sqrt{\frac{1+tan^2A}{1}-1}$

$=\sqrt{\frac{1+tan^2A-1}{1}}$

$=\sqrt{tan^2A}$

$=\boxed{tanA}$

= RHS

Hope it helps!

Answer 2

Answer: sorry bro it will take some time to solve this

But I Will be back to you

Step-by-step explanation:

Given, A + B = 90° (or) B = 90° - A.

LHS:

=\sqrt{\frac{tanAtanB+tanAcotB}{sinAsecB} - \frac{sin^2B}{cos^2A}}

=\sqrt{\frac{tanAtan(90-A)+tanAcot(90-A)}{sinAsec(90-A)}-\frac{sin^2(90-A)}{cos^2A}}

=\sqrt{\frac{tanAcotA+tanAtanA}{sinAcosecA}-\frac{cos^2A}{cos^2A}}

=\sqrt{\frac{tanA*\frac{1}{tanA}+tan^2A}{sinA*\frac{1}{sinA}}-\frac{cos^2A}{cos^2A}}

=\sqrt{\frac{1+tan^2A}{1}-1}

=\sqrt{\frac{1+tan^2A-1}{1}}

=\sqrt{tan^2A}

=\boxed{tanA}

= RHS

I don't know if it is correct or not but

Hope it helps