Prove that cosec²x(tan²x-sin²x)=tan²x

Answer 1

Starting with the left-hand side:
cosec^2x(tan^2x - sin^2x)

Using the identities:
cosec^2x = 1/sin^2x
tan^2x = sin^2x/cos^2x

We can rewrite the left-hand side as:
1/sin^2x * (sin^2x/cos^2x - sin^2x)

Simplifying:
1/sin^2x * [(sin^2x - sin^2x*cos^2x)/cos^2x]

Expanding the numerator:
1/sin^2x * [sin^2x(1 - cos^2x)/cos^2x]

Using the identity:
1 - cos^2x = sin^2x

1/sin^2x * [sin^2x*sin^2x/cos^2x]

Simplifying:
1/sin^2x * [sin^4x/cos^2x]

Simplifying the fraction:
1/sin^2x * tan^2x

Multiplying fractions:
tan^2x * 1/sin^2x

This simplifies to:
tan^2x/cos^2x

Using the identity tan^2x = sin^2x/cos^2x:
sin^2x/cos^2x

This is equal to the right-hand side, therefore proving that:
cosec^2x(tan^2x - sin^2x) = tan^2x.