Question

Prove trig identity
tan^2(x)-sin^2(x)=sin^2(x)tan^2(x)

Answer 1

Answer:

Step-by-step explanation:

taking LHS -

tan^2(x)-sin^2(x)

Break tan(x) into sin(x)/cos(x),

=[sin^2(x)/cos^2(x)]-sin^2(x)

taking LCM of denominator,

=[sin^2(x)-sin^2(x)cos^2(x)]/cos^2(x)

taking sin^2(x) common from numerator,

=[sin^2(x){1-cos^2(x)}]/cos^2(x)

and now we know that sin^2(x)/cos^2(x)=tan^2(x) and [1-cos^2(x)]=sin^2(x),so

=[sin^2(x)/cos^2(x)][1-cos^2(x)]

=tan^2(x)sin^2(x)

rearranging,

=sin^2(x)tan^2(x)

hence proved. and please make this a brainliest answer.