Question

prove the identity that (tan^2 x) - (sin^2 x) = (tan^2 x)(sin^2 x)

Answer 1

Replace with an equivalent expression using the fundamental identities.
((sin^(2)x)/(cos^(2)x))-(sin^(2)x)

Multiply -1 by each term inside the parentheses.
(sin^(2)x)/(cos^(2)x)-sin^(2)x

Combine (sinx^(2))/(cosx^(2))-sinx^(2) into a single expression by finding the least common denominator (LCD).
The LCD of (sinx^(2))/(cosx^(2))-sinx^(2) is cosx^(2).
(sin^(2)x-sin^(2)xcos^(2)x)/(cos^(2)x)

factor out the GCF of sinx^(2) from sinx^(2)-sinx^(2)cosx^(2).
(sin^(2)x(1-cos^(2)x))/(cos^(2)x)

Multiply sinx^(2)*sinx^(2) to get sinx^(4) in the numerator.
(sin^(4)x)/(cos^(2)x)

multiply -1 by each term inside the parentheses.
(sin^(2)x)/(cos^(2)x)-sin^(2)x

Combine (sinx^(2))/(cosx^(2))-sinx^(2) into a single expression by finding the least common denominator (LCD). The LCD of (sinx^(2))/(cosx^(2))-sinx^(2) is cosx^(2).
(=sin^(2)x-sin^(2)xcos^(2)x)/(cos^(2)x)

factor out the GCF of sinx^(2) from sinx^(2)-sinx^(2)cosx^(2).
=(sin^(2)x(1-cos^(2)x))/(cos^(2)x)

Multiply sinx^(2)*sinx^(2) to get sinx^(4) in the numerator.
=(sin^(4)x)/(cos^(2)x)

rewrite the expression using the (sinx)/(cosx)=tanx identity.
=sin^(2)xtan^(2)x

Answer 2

quite long, lol

Answer 3

whoh ok thanks

Answer 4

(sin(x))^2/(cos(x))^2 - (sin(x))^2
=>(sin(x))^2((1/(cos(x))^2)-1)

1/(cos(x))^2) = sec^2x
as we know (sec(x))^2-1) = (tan(x))^2
hence we have LHS

Answer 5

gracias por ayudandome