a. Express ^ 1x in terms of inverse of sin function. b. Prove by mathematical induction: 1 + 3 + 5 + ... + (2n 1) = n^2 c. Find the adjoint of the matrix pmatrix 2 & 5 \\ 3 & 7 pmatrix

a. Express ${\tan }^{-1}x$ in terms of inverse of sin function.

To express ${\tan }^{-1}x$ in terms of ${\sin }^{-1}x$, we use the following trigonometric identity:

$${\tan }^{-1}x={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$$

Derivation:

We know that: $$\tan \theta =x⟹\sin \theta =\frac{x}{\sqrt{1+x^2}}$$ where $\theta ={\tan }^{-1}x$.

Taking the inverse sine on both sides: $$\theta ={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$$

Thus, $${\tan }^{-1}x={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$$

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b. Prove by mathematical induction: $1+3+5+...+(2n-1)=n^2$

Base Case (n = 1): $$\text{LHS}=1=2(1)-1$$ $$\text{RHS}=1^2=1$$ Since LHS = RHS, the statement holds for $n=1$.

Inductive Step: Assume the statement is true for $n=k$, i.e., $$1+3+5+...+(2k-1)=k^2$$

Now, prove for $n=k+1$: $$1+3+5+...+(2k-1)+[2(k+1)-1]=(k+1{)}^2$$ $$k^2+(2k+1)=k^2+2k+1=(k+1{)}^2$$

Thus, the statement holds for $n=k+1$.

By the principle of mathematical induction, the statement is true for all positive integers $n$.

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c. Find the adjoint of the matrix $\left(\begin{array}{cc}2& 5\\ 3& -7\end{array}\right)$

For a $2\times 2$ matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the adjoint of $A$ is given by: $$\text{adj}(A)=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$$

Given: $$A=\left(\begin{array}{cc}2& 5\\ 3& -7\end{array}\right)$$

The adjoint of $A$ is: $$\text{adj}(A)=\left(\begin{array}{cc}-7& -5\\ -3& 2\end{array}\right)$$

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