a. Solve for general values of : + 2 + 3 = + 2 + 3 OR If a = 2 , b = 6 and c = 3 1 , find the angles of the triangle ABC. b. Prove that: vmatrix a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ac & bc &…

a. Solve for general values of $\theta$: $\sin \theta +\sin 2\theta +\sin 3\theta =\cos \theta +\cos 2\theta +\cos 3\theta$

Step 1: Simplify the left-hand side (LHS) and right-hand side (RHS) using trigonometric identities

We start by expressing $\sin 2\theta$ and $\cos 2\theta$ using double-angle identities: $$\sin 2\theta =2\sin \theta \cos \theta ,\cos 2\theta =2{\cos }^2\theta -1$$

Next, we use the sum-to-product identities for $\sin 3\theta$ and $\cos 3\theta$: $$\sin 3\theta =\sin (2\theta +\theta )=\sin 2\theta \cos \theta +\cos 2\theta \sin \theta$$ $$\cos 3\theta =\cos (2\theta +\theta )=\cos 2\theta \cos \theta -\sin 2\theta \sin \theta$$

However, a more efficient approach is to use the following identities for $\sin A+\sin B$ and $\cos A+\cos B$: $$\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$$ $$\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$$

Step 2: Group terms and apply identities

Rewrite the equation: $$\sin \theta +\sin 3\theta +\sin 2\theta =\cos \theta +\cos 3\theta +\cos 2\theta$$

Apply the sum-to-product identity to $\sin \theta +\sin 3\theta$: $$\sin \theta +\sin 3\theta =2\sin \left(\frac{\theta +3\theta }{2}\right)\cos \left(\frac{\theta -3\theta }{2}\right)=2\sin 2\theta \cos \theta$$

Similarly, apply the identity to $\cos \theta +\cos 3\theta$: $$\cos \theta +\cos 3\theta =2\cos 2\theta \cos \theta$$

Now, the equation becomes: $$2\sin 2\theta \cos \theta +\sin 2\theta =2\cos 2\theta \cos \theta +\cos 2\theta$$

Factor out $\sin 2\theta$ and $\cos 2\theta$: $$\sin 2\theta (2\cos \theta +1)=\cos 2\theta (2\cos \theta +1)$$

Step 3: Solve for $\theta$

Bring all terms to one side: $$\sin 2\theta (2\cos \theta +1)-\cos 2\theta (2\cos \theta +1)=0$$ $$(2\cos \theta +1)(\sin 2\theta -\cos 2\theta )=0$$

This gives two cases:

Case 1: $2\cos \theta +1=0$ $$\cos \theta =-\frac{1}{2}$$ $$\theta =2\pi n\pm \frac{2\pi }{3},n\in \mathbb{Z}$$

Case 2: $\sin 2\theta -\cos 2\theta =0$ $$\sin 2\theta =\cos 2\theta$$ $$\tan 2\theta =1$$ $$2\theta =\pi n+\frac{\pi }{4},n\in \mathbb{Z}$$ $$\theta =\frac{\pi n}{2}+\frac{\pi }{8},n\in \mathbb{Z}$$

Final Solution:

The general solutions are: $$\boxed{\theta =2\pi n\pm \frac{2\pi }{3}\text{or}\theta =\frac{\pi n}{2}+\frac{\pi }{8},n\in \mathbb{Z}}$$

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b. Prove that:

$$\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ac& bc& c^2+1\end{array}\right|=1+a^2+b^2+c^2$$

Step 1: Expand the determinant

Let $D$ be the given determinant: $$D=\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ac& bc& c^2+1\end{array}\right|$$

We expand $D$ using the first row: $$D=(a^2+1)\left|\begin{array}{cc}{b}^2+1& bc\\ bc& c^2+1\end{array}\right|-ab\left|\begin{array}{cc}ab& bc\\ ac& c^2+1\end{array}\right|+ac\left|\begin{array}{cc}ab& b^2+1\\ ac& bc\end{array}\right|$$

Step 2: Compute the 2x2 determinants

Compute each minor: $$\left|\begin{array}{cc}{b}^2+1& bc\\ bc& c^2+1\end{array}\right|=(b^2+1)(c^2+1)-(bc{)}^2=b^2{c}^2+b^2+c^2+1-b^2{c}^2=b^2+c^2+1$$

$$\left|\begin{array}{cc}ab& bc\\ ac& c^2+1\end{array}\right|=ab(c^2+1)-bc(ac)=ab{c}^2+ab-ab{c}^2=ab$$

$$\left|\begin{array}{cc}ab& b^2+1\\ ac& bc\end{array}\right|=ab(bc)-ac(b^2+1)=a{b}^{2}c-ab{c}^2-a{b}^2=a{b}^{2}c-ab{c}^2-a{b}^2$$

Step 3: Substitute back and simplify

Substitute the minors into the expansion: $$D=(a^2+1)(b^2+c^2+1)-ab(ab)+ac(a{b}^{2}c-ab{c}^2-a{b}^2)$$

Simplify each term: $$D=(a^2+1)(b^2+c^2+1)-a^2{b}^2+a^2{b}^2{c}^2-a^2{b}^2{c}^2-a^2{b}^{2}c$$

Notice that $a^2{b}^2{c}^2$ cancels out: $$D=(a^2+1)(b^2+c^2+1)-a^2{b}^2-a^2{b}^{2}c$$

Now, expand $(a^2+1)(b^2+c^2+1)$: $$=a^2{b}^2+a^2{c}^2+a^2+b^2+c^2+1$$

Substitute back: $$D=a^2{b}^2+a^2{c}^2+a^2+b^2+c^2+1-a^2{b}^2-a^2{b}^{2}c$$

Simplify further: $$D=a^2{c}^2+a^2+b^2+c^2+1-a^2{b}^{2}c$$

However, this does not match the desired result. Let's try an alternative approach using row operations or column operations.

Alternative Approach: Add and Subtract Rows

Let $D$ be the original determinant. We perform row operations to simplify it.

Step 1: Subtract $a$ times the first row from the second row, and $b$ times the first row from the third row.

But a more efficient method is to recognize that the determinant can be rewritten as: $$D=\left|\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right|+\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$$

The second determinant is simply 1. The first determinant is: $$\left|\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right|=a^2(b^2{c}^2-b^2{c}^2)-ab(ab{c}^2-bcac)+ac(abbc-b^{2}ac)$$ $$=0-ab(ab{c}^2-ab{c}^2)+ac(a{b}^{2}c-a{b}^{2}c)$$ $$=-a^2{b}^2{c}^2+a^2{b}^2{c}^2+0=0$$

This is incorrect. Let's use a different strategy.

Correct Approach: Use the identity for determinants of the form $A+I$

Notice that the matrix can be written as: $$\left[\begin{array}{ccc}{a}^2& ab& ac\\ ab& b^2& bc\\ ac& bc& c^2\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$

Let $M={\left[\begin{array}{ccc}a& b& c\end{array}\right]}^T\left[\begin{array}{ccc}a& b& c\end{array}\right]$, so the matrix is $M+I$.

The determinant of $M+I$ can be computed using the formula for rank-one updates: $$\det (M+I)=\det (I)+\text{tr}(\text{adj}(I)M)+\det (M)$$

However, a simpler approach is to recognize that: $$\det (M+I)=1+a^2+b^2+c^2$$

This is because the matrix $M$ has rank 1, and its eigenvalues are $a^2+b^2+c^2$, $0$, and $0$. Thus, the eigenvalues of $M+I$ are $a^2+b^2+c^2+1$, $1$, and $1$. The determinant is the product of the eigenvalues: $$\det (M+I)=(a^2+b^2+c^2+1)\cdot 1\cdot 1=1+a^2+b^2+c^2$$

Final Answer:

$$\boxed{1+a^2+b^2+c^2}$$