Chapter 5: Trigonometric Functions – Exercise 5.2

Trigonometric Functions – Exercise 5.2 – Q.1

we have,

cosec²θ - cot²θ = 1

⟹ cosec² θ = 1 + cot² θ

In the third quadrant cosec θ is negative

We have,

sin²θ + cos²θ = 1

⟹ sin²θ = 1 - cos²θ 

In the 2nd quadrant sin θ is positive and tan θ is negative

In the third quadrant cosec θ is negative

We have,

sin²θ + cos² θ = 1

⟹ cos² θ = 1 - sin² θ

In the 1^st quadrant cos θ is positive and tan θ is also positive

 

Trigonometric Functions – Exercise 5.2 – Q.2

We have,

sin²θ + cos²θ = 1

⟹ cos²θ = 1 - sin²θ

In the 2^st quadrant cos θ is negative and tan θ is also negative

 

Trigonometric Functions – Exercise 5.2 – Q.3

We have,

⟹ θ lies in the second quadrant and Φ lies in the third quadrant.

Now, sin² θ + cos² θ = 1

⟹ cos² θ = 1 - sin² θ

In the 2^st quadrant cos θ is negative and tan θ is also negative

In the third quadrant sec Φ is negative