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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 1st 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 2st 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 2st quadrant cos θ is negative and tan θ is also negative In the third quadrant sec Φ is negative
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
sin2θ + cos2 θ = 1
⟹ cos2 θ = 1 - sin2 θ
In the 1st quadrant cos θ is positive and tan θ is also positive
⟹ cos2θ = 1 - sin2θ
In the 2st quadrant cos θ is negative and tan θ is also negative
⟹ θ lies in the second quadrant and Φ lies in the third quadrant.
Now, sin2 θ + cos2 θ = 1
In the third quadrant sec Φ is negative
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Chapter 5: Trigonometric Functions –...
