trigonometric help please.......10th grade?

Question

we have to prove LHS = RHS please help
LHS:
Tan squared A - Tan squared B
RHS:
Sin squared A-Sin squared B/the whole thing upon Cos squared A- Cos squared B

sorry its not cos squared A - cos squared B IN THE DENOMINATOR OF RHS IT IS COS SQUARED A MULTIPLIED BY COS SQUARED B

Answer 1

Question rewritten:

show that
tan^2 A - tan^2 B = (sin^2 A - sin^2 B) / (cos^2 A cos^2 B).

Rewrite the left side using tan x = sin x / cos x:

sin^2 A / cos^2 A - sin^2 B / cos^2 B

Put everything over the common denominator cos^2 A cos^2 B:

(cos^2 B/cos^2 B) (sin^2 A/cos^2 A) - (sin^2 B/cos^2 B) (cos^2 A/cos^2 A) =
[cos^2 B sin^2 A - sin^2 B cos^2 A] / [cos^2 A cos^2 B] =
[cos^2 B sin^2 A + sin^2 B sin^2 A - sin^2 B sin^2 A - sin^2 B cos^2 A] / [cos^2 A cos^2 B] =
[sin^2 A[cos^2 B + sin^2 B] - sin^2 B[sin^2 A+cos^2 A]] / [cos^2 A cos^2 B] =
[sin^2 A - sin^2 B] / [cos^2 A cos^2 B]

In the last step, you use the identity cos^2 B + sin^2 B = 1.

(I know several people have answered this already, but I saw this question just as I was about to go out, and had to address it once I returned, regardless :)

Answer 2

LHS= tan^2 A - tan^2 B
LHS= {sin^2 A/ cos^2 A} - {sin^2 B/ cos^2 B}
= {sin^2 A*cos^2 B - sin^2 B cos^2 A}/{cos^2 A cos^2 B}
= {sin^2 A*(1-sin^2 B) - sin^2 B (1- sin^2 A)}/{cos^2 A cos^2 B}
= {sin^2 A -sin^2 A*sin^2 B - sin^2 B +sin^2 A*sin^2 B)}/{cos^2 A cos^2 B}
{sin^2 A - sin^2 B }/{cos^2 A cos^2 B}=RHS

Answer 3

LHS= tan^2 A - tan^2 B
LHS= {sin^2 A/ cos^2 A} - {sin^2 B/ cos^2 B}
= {sin^2 A*cos^2 B - sin^2 B cos^2 A}/{cos^2 A cos^2 B}
= {sin^2 A*(1-sin^2 B) - sin^2 B (1- sin^2 A)}/{cos^2 A cos^2 B}
= {sin^2 A -sin^2 A*sin^2 B - sin^2 B +sin^2 A*sin^2 B)}/{cos^2 A cos^2 B}
{sin^2 A - sin^2 B }/{cos^2 A cos^2 B}=RHS

Answer 4

There is something wrong with your question. As stated, the LHS and RHS are not equal. What you meant to write was:

(tan(a)^2 - tan(b)^2) = (sin(a)^2 - sin(b)^2)/(cos(a)^2 * cos(b)^2)

Notice the mulitplication in the denominator.

All you really need to know that sin(x)^2 + cos(x)^2 = 1 and that tan(x) = sin(x)/cos(x). Convert bothsides to all cosines and play around a little and it should work out.

Answer 5

sin^ A over cos^2 A is the same as saying tan^2 A.

sin over cos = tan. so just work with sin's and cos.

sin^2 A / cos^2 A - sin^2 B / cos^2 B correct?

sin^2 A / cos^2 A = tan^2 A.

sin^2 B / cos^2 B = tan^2 B.

so you're left with tan^2 A - tan^2 B.

tada :D just work with sin's and cos's to convert it to tan's.

Answer 6

LHS is not equal to RHS . (the question is wrong) . lets try a simple test .

if LHS = RHS . It is true for all values of A and B . so substitute A=60 deg and B= 30 deg . LHS = 8/3 and RHS = -1 . So, LHS is not equal to RHS

Answer 7

LHS:
sin squareA/cos square A - sin square B/cos square B

after this ur on ur own.

Answer 8

Is the RHS correct?
this is what I get for LHS

LHS = tan^2(A) - tan^2(B)
= sin^2(A)/cos^2(A) - sin^2(B)/cos^2(B)
= (sin^2(A)cos^(B) - cos^(A)sin^(B))/cos^2(A)cos^2(B)
use cos^2(B) = 1 - sin^2(B) and use cos^2(A) = 1 - sin^2(A)
= (sin^2(A) - sin^2(B))/(cos^2(A)cos^2(B))

check your RHS

hmmm random values too reveal the incorrectness of your question.

Answer 9

Expand the LHS into sines and cosines and find the least common denominator

Answer 10

Wow... you're in 10th grade and doing trigonometry? Most peope I know in 10th grade are doing Algebra 1 and 2!