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Let $E = K(a,b). $Suppose gcd of degree of minimal polynomial of $a$ and $b$ over $K$ is $1$, then does it imply that the degree of the product of the minimal polynomials is same as $[E:K]$?

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    Yes, if you meant that $[E:K]$ is the product of the degrees of the minimal polynomials of $a$ resp. $b$. The way you wrote it makes it look like you mean the degree of $m_a(x)m_b(x)$ which is the **sum** of the degrees $\deg m_a+\deg m_b$, but that can't be... – 2017-01-31
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    if deg are prime numbers – 2017-01-31
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    Consider the example case of $a=\root3\of2$, $b=\sqrt2$ and $K=\Bbb{Q}$. Then $E=\Bbb{Q}(a,b)=\Bbb{Q}(\root6\of2)$ is a degree six extension. The product of minimal polynomials $(x^3-2)(x^2-2)$ on the other hand has degree five. Here, by the formula for the degree of a tower of extensions, $[E:\Bbb{Q}]$ must be a multiple of $3=[\Bbb{Q}(a):\Bbb{Q}]$ and $2=[\Bbb{Q}(\sqrt2):\Bbb{Q}]$, so it must be a multiple of six as explained in MooS's answer. – 2017-01-31

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Yes. Let $m=[K(a):K]$ and $n=[K(b):K]$ be the degrees of the minimal polynomials. Then $[E:K]$ is divisible by $m$ and $n$ by the tower law. If $m$ and $n$ are co-prime, we get that $[E:K]$ is divisible $mn$, hence $mn \leq [E:K]$. The converse inequality holds trivially without any further assumptions.