Question

please tell this i will mark u as a brainlist

Answer 1

Answer is in photo
I have not solved by mobile because it is too long
Mark it as brainliest bro.

Answer 2

Step-by-step explanation:

(i)

$Given:tan^2A-tanB$

$=\frac{sin^2A}{cos^2A}-\frac{sin^2B}{cos^2B}$

$=\frac{sin^2Acos^2B-sin^2BCos^2A}{cos^2Acos^2B}$

$=\frac{sin^2A(1-sin^2B)-sin^2B(1-sin^2A)}{cos^2Acos^2B}$

$=\frac{sin^2A-sin^2Asin^2B-sin^2B+sin^2Bsin^2A}{cos^2Acos^2B}$

$=\frac{sin^2A-sin^2B}{cos^2Acos^2B}$

(ii)

$Given:\frac{sin^2A-sin^2B}{cos^2Acos^2B}$

$=\frac{(1-cos^2A)-(1-cos^2B)}{cos^2Acos^2B}$

$=\frac{1-cos^2A-1+cos^2B}{cos^2Acos^2B}$

$=\frac{cos^2B-cos^2A}{cos^2Acos^2B}$

∴ If 1 = 2 = 3, then 1 = 3, 2 = 3.

Hope it helps!