Question

Tan^2A-sin^2A= tan^2Asin^2A

Answer 1

$tan^2A-sin^2A=tan^2Asin^2A$

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we have :-

LEFT HAND SIDE :-

$=tan^2A-sin^2A\\ \\ \\=\dfrac{sin^2A}{cos^2A}-sin^2A\\ \\ \\=sin^2A(\dfrac{1}{cos^2A}-1)\\ \\ \\=sin^2A(sec^2A-1)\\ \\ \\=sin^2A*tan^2A\\ \\ \\=RHS$

$\rule{200}{2}$

TRIGNOMETRIC IDENTITIES:-

$1) sin^2x+cos^2x=1\\ \\ \\ \\2)1+tan^2x=sec^2x\\ \\ \\ \\3)1+cot^2x=cosec^2x$

$\rule{200}{2}$

Hence,

LEFT HAND SIDE = RIGHT HAND SIDE

Thus,

Proved!

Answer 2

Answer:

Trigonometry

Refer the attached picture.