Find the values of other five trigonometric functions in Exercises 1 to 5.
Question 1. , x lies in third quadrant.
Solution.

Question 2. , x lies in second quadrant.
Solution.

Question 3. , xlies in third quadrant.
Solution.

Question 4. , x lies in fourth quadrant.
Solution.

Question 5. , x lies in second quadrant.
Solution.
It is given that
tan x = – 5/12
We can write it as

We know that
1 + tan2 x = sec2 x
We can write it as
1 + (-5/12)2 = sec2 x
Substituting the values
1 + 25/144 = sec2 x
sec2 x = 169/144
sec x = ± 13/12
Here x lies in the second quadrant so the value of sec x will be negative
sec x = – 13/12
We can write it as

Find the values of the trigonometric functions in Exercises 6 to 10.
Question 6. sin 765°
Solution.

Question 7. cosec (-1410°)
Solution.

Question 8.
Solution.

Question 9.
Solution.

Question 10.
Solution.
