Question

Verify the following identity:
Tan2 x/ 1+ tan2 x = sin2 x

Answer 1

Someone answer me please?!?:)

Answer 2

\[\frac{\tan^2x}{1 + \tan^2x} = \sin^2x\]

Answer 3

Yes that is the question but I need to find the identity and how I did it.. Make one side equal the other .. Idk how to do that with this one problem

Answer 4

Well, basically since the right side is just a single term, we'll need to manipulate the left side so that it is also one term. The easiest way to do that is to turn \(1 + \tan^2x\) into an expression that is just one term. We can do that by expressing it like so:

Answer 5

As you know: \(\tan^2x = \frac{\sin^2x}{\cos^2x}\)

So we can re-write the left hand side as:
\[\dfrac{\tan^2x}{1 + \dfrac{\sin^2x}{\cos^2x}}\]

Answer 6

Now we can convert 1 to \(\dfrac{\cos^2x}{\cos^2x}\):

\[\dfrac{\tan^2x}{\dfrac{\cos^2x}{\cos^2x} + \dfrac{\sin^2x}{\cos^2x}}\]

Answer 7

Then combine fractions in the denominator to get:

\[\dfrac{\tan^2x}{\dfrac{\cos^2x +\sin^2x}{\cos^2x}}\]

Notice that \(\cos^2x + \sin^2x = 1\) so

\[\dfrac{\tan^2x}{\dfrac{1}{\cos^2x}}\]

Answer 8

Next express the fraction in the following form:

\(\tan^2x \div \dfrac{1}{\cos^2x}\)

And use the reciprocal rule to express it as:

\(\tan^2x \times\cos^2x\)

Express \(\tan^2x\) in explicit form:

\(\dfrac{\sin^2x}{\cos^2x} \times \cos^2x\)

And I'll leave the rest to you

Answer 9

Ok thanks a lot!!!

Answer 10

You're welcome.