a) In any triangle ABC, prove that 12(B C) = b cb + c A2 . [4] OR Solve: x x = 2 . b) Show that: vmatrix 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 vmatrix = (y z)(z x)(x y) . [4]

a) Proof: $\tan \frac{1}{2}(B-C)=\frac{b-c}{b+c}\cot \frac{A}{2}$

We start with the Law of Sines in triangle $ABC$: $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R,$$ where $R$ is the circumradius. Thus, we can express the sides as: $$b=2R\sin B,c=2R\sin C.$$

Step 1: Express $\tan \frac{1}{2}(B-C)$

Using the tangent of a difference formula: $$\tan \frac{1}{2}(B-C)=\frac{\tan \frac{B}{2}-\tan \frac{C}{2}}{1+\tan \frac{B}{2}\tan \frac{C}{2}}.$$

Step 2: Use half-angle identities

Recall the half-angle formula for tangent: $$\tan \frac{\theta }{2}=\frac{1-\cos \theta }{\sin \theta }=\frac{\sin \theta }{1+\cos \theta }.$$ We also know that: $$\cot \frac{A}{2}=\frac{1+\cos A}{\sin A}.$$

Step 3: Express $\frac{b-c}{b+c}$

Substitute $b=2R\sin B$ and $c=2R\sin C$: $$\frac{b-c}{b+c}=\frac{\sin B-\sin C}{\sin B+\sin C}.$$ Using the sum-to-product identities: $$\sin B-\sin C=2\cos \frac{B+C}{2}\sin \frac{B-C}{2},$$ $$\sin B+\sin C=2\sin \frac{B+C}{2}\cos \frac{B-C}{2}.$$ Thus: $$\frac{b-c}{b+c}=\frac{\cos \frac{B+C}{2}\sin \frac{B-C}{2}}{\sin \frac{B+C}{2}\cos \frac{B-C}{2}}=\cot \frac{B+C}{2}\tan \frac{B-C}{2}.$$

Step 4: Simplify using angle sum in a triangle

Since $A+B+C=\pi$, we have: $$\frac{B+C}{2}=\frac{\pi -A}{2}=\frac{\pi }{2}-\frac{A}{2}.$$ Thus: $$\cot \frac{B+C}{2}=\cot \left(\frac{\pi }{2}-\frac{A}{2}\right)=\tan \frac{A}{2}.$$ Substituting back: $$\frac{b-c}{b+c}=\tan \frac{A}{2}\tan \frac{B-C}{2}.$$

Step 5: Combine with $\cot \frac{A}{2}$

Multiply both sides by $\cot \frac{A}{2}$: $$\frac{b-c}{b+c}\cot \frac{A}{2}=\tan \frac{A}{2}\tan \frac{B-C}{2}\cot \frac{A}{2}=\tan \frac{B-C}{2}.$$ This proves the required identity: $$\tan \frac{1}{2}(B-C)=\frac{b-c}{b+c}\cot \frac{A}{2}.$$

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OR

Solution: $\sec x\tan x=\sqrt{2}$

Step 1: Rewrite using trigonometric identities

Recall that: $$\sec x\tan x=\frac{\sin x}{{\cos }^{2}x}.$$ Thus, the equation becomes: $$\frac{\sin x}{{\cos }^{2}x}=\sqrt{2}.$$

Step 2: Use substitution

Let $t=\tan x$. Then: $$\sin x=\frac{t}{\sqrt{1+t^2}},{\cos }^{2}x=\frac{1}{1+t^2}.$$ Substituting: $$\frac{\frac{t}{\sqrt{1+t^2}}}{\frac{1}{1+t^2}}=\sqrt{2}⟹t(1+t^2)=\sqrt{2}\sqrt{1+t^2}.$$ Square both sides: $$t^2(1+t^2{)}^2=2(1+t^2).$$ Let $u=1+t^2$: $$t^2{u}^2=2u⟹u(t^{2}u-2)=0.$$ Since $u=1+t^2\ne 0$, we have: $$t^{2}u=2⟹t^2(1+t^2)=2⟹t^4+t^2-2=0.$$ Let $v=t^2$: $$v^2+v-2=0⟹(v+2)(v-1)=0.$$ Thus, $v=1$ (since $v=-2$ is invalid). Therefore: $$t^2=1⟹t=\pm 1.$$

Step 3: Solve for $x$

If $t=\tan x=1$, then: $$x=\frac{\pi }{4}+n\pi ,n\in \mathbb{Z}.$$ If $t=\tan x=-1$, then: $$x=-\frac{\pi }{4}+n\pi ,n\in \mathbb{Z}.$$

Verification

Check $x=\frac{\pi }{4}$: $$\sec \frac{\pi }{4}\tan \frac{\pi }{4}=\sqrt{2}\cdot 1=\sqrt{2}.$$ Similarly, $x=-\frac{\pi }{4}$ also satisfies the equation.

Final Answer: $$\boxed{x=\frac{\pi }{4}+n\pi \text{or}x=-\frac{\pi }{4}+n\pi ,n\in \mathbb{Z}}$$

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b) Proof: $\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=(y-z)(z-x)(x-y)$

Step 1: Expand the determinant

The determinant of a $3\times 3$ matrix: $$\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=1\cdot (y{z}^2-y^{2}z)-x(z^2-y^2)+x^2(z-y).$$

Step 2: Simplify each term

$$=y{z}^2-y^{2}z-x{z}^2+x{y}^2+x^{2}z-x^{2}y.$$

Step 3: Factorize

Group terms: $$=(y{z}^2-x{z}^2)+(x{y}^2-y^{2}z)+(x^{2}z-x^{2}y).$$ Factor out common terms: $$=z^2(y-x)+y^2(x-z)+x^2(z-y).$$

Step 4: Rewrite using differences

Notice that: $$(y-x)=-(x-y),(x-z)=(x-y)+(y-z),(z-y)=-(y-z).$$ Substitute and simplify: $$=-z^2(x-y)+y^2(x-y+y-z)-x^2(y-z).$$ $$=(x-y)(-z^2+y^2)+y^2(y-z)-x^2(y-z).$$ $$=(x-y)(y^2-z^2)+(y-z)(y^2-x^2).$$ Factor further: $$=(x-y)(y-z)(y+z)+(y-z)(y-x)(y+x).$$ $$=(y-z)(x-y)(y+z-y-x)=(y-z)(x-y)(z-x).$$ Rearrange: $$=(y-z)(z-x)(x-y).$$

Final Answer: $$\boxed{(y-z)(z-x)(x-y)}$$