Sure! Let's solve each part of the question step by step.
### Part (i)
We need to find the value of [tex]\(\sin^2 45^\circ + \cos^2 60^\circ\)[/tex].
First, let's find [tex]\(\sin 45^\circ\)[/tex] and [tex]\(\cos 60^\circ\)[/tex]:
[tex]\[
\sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 60^\circ = \frac{1}{2}
\][/tex]
Now square these values:
[tex]\[
\sin^2 45^\circ = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2}, \quad \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4}
\][/tex]
Add these squares together:
[tex]\[
\sin^2 45^\circ + \cos^2 60^\circ = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}
\][/tex]
So, the exact value of [tex]\(\sin^2 45^\circ + \cos^2 60^\circ\)[/tex] is [tex]\(\frac{3}{4}\)[/tex].
### Part (ii)
We need to find the value of [tex]\(\sin \frac{\pi}{3} + \cos \frac{\pi}{6}\)[/tex].
First, let's find [tex]\(\sin \frac{\pi}{3}\)[/tex] and [tex]\(\cos \frac{\pi}{6}\)[/tex]:
[tex]\[
\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}
\][/tex]
Add these values together:
[tex]\[
\sin \frac{\pi}{3} + \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}
\][/tex]
So, the exact value of [tex]\(\sin \frac{\pi}{3} + \cos \frac{\pi}{6}\)[/tex] is [tex]\(\sqrt{3}\)[/tex].
### Part (iii)
We need to find the value of [tex]\(\csc \frac{\pi}{6} + \sec \frac{\pi}{3}\)[/tex].
First, let's find [tex]\(\csc \frac{\pi}{6}\)[/tex] and [tex]\(\sec \frac{\pi}{3}\)[/tex]:
[tex]\[
\csc \frac{\pi}{6} = \frac{1}{\sin \frac{\pi}{6}}, \quad \sec \frac{\pi}{3} = \frac{1}{\cos \frac{\pi}{3}}
\][/tex]
[tex]\[
\sin \frac{\pi}{6} = \frac{1}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}
\][/tex]
[tex]\[
\csc \frac{\pi}{6} = \frac{1}{\frac{1}{2}} = 2, \quad \sec \frac{\pi}{3} = \frac{1}{\frac{1}{2}} = 2
\][/tex]
Add these values together:
[tex]\[
\csc \frac{\pi}{6} + \sec \frac{\pi}{3} = 2 + 2 = 4
\][/tex]
So, the exact value of [tex]\(\csc \frac{\pi}{6} + \sec \frac{\pi}{3}\)[/tex] is [tex]\(4\)[/tex].
### Proof
We need to show that [tex]\(\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = 1\)[/tex].
First, let's find the values involved:
[tex]\[
\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \sin 30^\circ = \frac{1}{2}
\][/tex]
Now, use these values to calculate the expression:
[tex]\[
\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = \left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2} \cdot \frac{1}{2}\right)
\][/tex]
[tex]\[
= \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1
\][/tex]
Thus, [tex]\(\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = 1\)[/tex] is indeed true.
So, we've verified this identity successfully.
