Question

plz solve it.. correct answer will be marked as brainliest and his or her account will be followed...promise...​

Answer 1

Answer:

Photo Attached above...

Step-by-step explanation:

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Answer 2

To Prove:

$tan^2A-sin^2A=tan^2A.sin^2A\\LHS= tan^2A-sin^2A\\=\frac{sin^2A}{cos^2A}-sin^2A\\ \\=\frac{sin^2A-cos^2Asin^2A}{cos^2A}\\ \\=\frac{sin^2A(1-cos^2A)}{cos^2A}\\ \\=\frac{sin^2A}{cos^2A}(1-cos^2A)\\ \\=tan^2A(1-cos^2A)\\=tan^2A.sin^2A$

Hence, LHS = RHS.