Help please I don't understand where to get the degrees from. And what degrees are each
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SOLUTION:
(i)
[tex]\begin{gathered} \sin \text{ }\theta\text{ = }\cos \text{ }\theta \\ \end{gathered}[/tex]Sine of an angle is complementary to the Cosine of an angle so for both to be the same the angle 90 degrees must be divided into.
[tex]\theta=45^0[/tex](ii)
[tex]\begin{gathered} \tan \text{ }\theta\text{ = 0} \\ \theta\text{ = }\tan ^{-1}(0) \\ \theta\text{ = 0} \end{gathered}[/tex](iii)
[tex]\csc \text{ }\theta\text{ > sec }\theta[/tex]As said earlier sine and cosine of angles are complementary so cosecant and secant of angles are also complementary.
[tex]\begin{gathered} \text{When }\theta\text{ = 30} \\ \csc \text{ }\theta\text{ > sec }\theta \\ \frac{1}{\sin30}\text{ > }\frac{1}{\cos30} \\ \\ 2\text{ > 1.15} \\ \\ \csc \text{ }\theta\text{ > sec }\theta\text{ when }\theta=30^0 \end{gathered}[/tex]