a) Consider ^ 1 ( 12 ) + ^ 1(3) = ^ 1 x . What is the value of x ? [3] b) Show that 12 [( a b) ( a + b)] = a b [2]

a) Solution for ${\tan }^{-1}\left(\frac{1}{2}\right)+{\cot }^{-1}(3)={\tan }^{-1}x$

We are given: $${\tan }^{-1}\left(\frac{1}{2}\right)+{\cot }^{-1}(3)={\tan }^{-1}x$$

Step 1: Recall that ${\cot }^{-1}(y)={\tan }^{-1}\left(\frac{1}{y}\right)$. Thus, $${\cot }^{-1}(3)={\tan }^{-1}\left(\frac{1}{3}\right)$$

Step 2: Let $$A={\tan }^{-1}\left(\frac{1}{2}\right)\text{and}B={\tan }^{-1}\left(\frac{1}{3}\right)$$ We need to find $\tan (A+B)$.

Using the tangent addition formula: $$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$

Step 3: Substitute $\tan A=\frac{1}{2}$ and $\tan B=\frac{1}{3}$: $$\tan (A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot \frac{1}{3}}=\frac{\frac{5}{6}}{1-\frac{1}{6}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1$$

Step 4: Since $\tan (A+B)=1$, we have: $$A+B={\tan }^{-1}(1)⟹{\tan }^{-1}x={\tan }^{-1}(1)$$

Thus, the value of $x$ is: $$\boxed{1}$$

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b) Proof that $\left|\frac{1}{2}[(\vec{a}-\vec{b})\times (\vec{a}+\vec{b})]\right|=|\vec{a}||\vec{b}|\sin \theta$

Step 1: Expand the cross product inside the absolute value: $$(\vec{a}-\vec{b})\times (\vec{a}+\vec{b})=\vec{a}\times \vec{a}+\vec{a}\times \vec{b}-\vec{b}\times \vec{a}-\vec{b}\times \vec{b}$$

Step 2: Simplify using properties of cross products: $$\vec{a}\times \vec{a}=\vec{0}\text{and}\vec{b}\times \vec{b}=\vec{0}$$ Also, $\vec{b}\times \vec{a}=-(\vec{a}\times \vec{b})$. Thus: $$(\vec{a}-\vec{b})\times (\vec{a}+\vec{b})=\vec{a}\times \vec{b}-(-\vec{a}\times \vec{b})=2(\vec{a}\times \vec{b})$$

Step 3: Substitute back into the original expression: $$\left|\frac{1}{2}[2(\vec{a}\times \vec{b})]\right|=|\vec{a}\times \vec{b}|$$

Step 4: Recall that the magnitude of the cross product is: $$|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Thus, we have shown: $$\left|\frac{1}{2}[(\vec{a}-\vec{b})\times (\vec{a}+\vec{b})]\right|=|\vec{a}||\vec{b}|\sin \theta$$