Proof of Multiple angle trigonometri ratios If $\tan^2{A}=1+2\tan^2B$ then prove that $\cos2B=1+2\cos2A$
I could not relate from $\tan$ to $\cos$
 A: Add $1$ to both sides of the tangent condition.
We get $\tan^2A+1=2+2\tan^2B$.
Simplifying with the identity $\tan^2\theta+1=\sec^2\theta$, we have:
$$\sec^2A=2\sec^2B$$
Taking the reciprocals of both sides and multiplying by $4$, we get:
$$4\cos^2A=2\cos^2B$$
Remembering the identity $2\cos^2\theta-1=\cos2\theta$, we can subtract $1$ from each side to get:
$$2(2\cos^2A-1)+1=2\cos^2B-1$$
which simplifies to:
$$2\cos2A+1=\cos2B$$
A: Notice, given that $$\tan^2A=1+2\tan^2B$$
$$\tan^2A+1=2+2\tan^2B$$
$$1+\tan^2A=2(1+\tan^2B)$$
$$\frac{1}{1+\tan^2B}=\frac{2}{1+\tan^2A}$$
$$\frac{1-\tan^2 B}{1+\tan^2B}=\frac{2(1-\tan^2 B)}{1+\tan^2A}$$
$$\cos 2B=\frac{2-2\tan^2B}{1+\tan^2A}$$
setting $2\tan^2B=\tan^2A-1, $
$$\cos 2B=\frac{2-\tan^2A+1}{1+\tan^2A}$$
$$\cos 2B=\frac{(1+\tan^2A)+2(1-\tan^2A)}{1+\tan^2A}$$
$$\cos 2B=1+2\frac{(1-\tan^2A)}{1+\tan^2A}$$
$$\cos 2B=1+2\cos 2A$$
A: One may observe that, in general,
$$
\frac1{\cos^2 a}=\frac{\cos^2 a+\sin^2 a}{\cos^2 a}=1+\tan^2 a \tag1
$$ and
$$
\cos^2 a=\frac{1+\cos (2a)}2 \tag2.
$$ Thus from
$$
\tan^2{A}=1+2\tan^2B
$$ we get
$$
1+\tan^2{A}=2+2\tan^2B
$$ or using $(1)$
$$
\frac1{\cos^2 A}=\frac2{\cos^2 B}
$$ and using $(2)$
$$
\frac2{1+\cos (2A)}=\frac4{1+\cos (2B)}
$$ giving
$$
1+\cos (2B)=2+2\cos (2A)
$$ that is
$$
\cos (2B)=1+2\cos (2A)
$$ as announced.
A: $$\frac{\sin^2 A}{\cos^2A}=1+2\frac{\sin^2 B}{\cos^2B}$$
using the identity $$\tan A = \frac{\sin A}{\cos A}$$
now using $$\cos^2A+\sin^2A = 1$$
$$\frac{1-\cos^2 A}{\cos^2 A} = 1+ 2\frac{1-\cos^2 B}{\cos^2 B}$$
$$\frac{1}{\cos^2 A}-1 = 1+ \frac{2}{\cos^2 B} -2$$
$$2\cos^2 A=\cos^2 B$$
and using the identity $$\cos^2 A = \frac{1+ \cos 2A}{2}$$
$$1+\cos 2A = \frac{1+\cos 2B}{2}$$
and
$$1+2\cos 2A = \cos 2B$$
