Question

How do you prove `(tan^2x+1)(cos^2x+1)=tan^2x+2`?

Answer 1

I would rewrite everything in terms of sine and cosine.

`(sin^2x/cos^2x + 1)(cos^2x + 1) = sin^2x/cos^2x+ 2`

`((sin^2x + cos^2x)/cos^2x)(cos^2x +1) = sin^2x/cos^2x +2`

`1/cos^2x(cos^2x + 1) = sin^2x/cos^2x+ 2`

`cos^2x/cos^2x+ 1/cos^2x = sin^2x/cos^2x+ 2`

`1 + 1/cos^2x = sin^2x/cos^2x+ 2`

Now we rewrite the right hand side as follows.

`1+ 1/cos^2x = (sin^2x + 2cos^2x)/cos^2x`

`1 + 1/cos^2x= (sin^2x+ cos^2x+ cos^2x)/cos^2x`

`1 + 1/cos^2x = (1 + cos^2x)/cos^2x`

`1 +1/cos^2x= 1/cos^2x + 1`

`LHS = RHS`

As required.

Hopefully this helps!

Answer 2

To prove an identity one should use identities and axioms to change only one side of the equation until it is identical to the other side.

Explanation:

Given: `(tan^2(x)+1)(cos^2(x)+1)=tan^2(x)+2`

Substitute `tan^2(x)+1= sec^2(x)`

`(sec^2(x))(cos^2(x)+1)=tan^2(x)+2`

Perform the multiplication:

`sec^2(x)cos^2(x)+sec^2(x)=tan^2(x)+2`

Use the identity `sec(x)=1/cos(x)`:

`1/cos^2(x)cos^2(x)+sec^2(x)=tan^2(x)+2`

The first term becomes 1:

`1+sec^2(x)=tan^2(x)+2`

Use the identity `sec^2(x) = tan^2(x)+1`

`1+tan^2(x)+1=tan^2(x)+2`

Combine like terms on the left:

`tan^2(x)+2=tan^2(x)+2` Q.E.D.