Answer:
See Explaination
Step-by-step explanation:
According To Question
- In Quadrant-1 (0* to 90*) ( All the Trignometry Functions are Positive )
- In Quadrant-2 (90* to 180*) (Only Sine & Cosecant are Positive)
- In Quadrant-3 (180* to 270*) (Only Tangent & Cotangent Are Positive)
- In Quadrant-4 (270* to 360*) ( Only Cosine & Secant Are Positive)
(Diagram, Please Find In Attachment)
Question 38 . Cos ∅ = 12/13 ∅ in Quadrant-4(Where only Cos & Sec are Positive all other Functions Are Negative in this Quadrant)
Now ∅ Represent a Right Angle Triangle with Base=12 Perpendicular=5 & Hypothenese=13
- other Five Trignometry Functions Are as Follow
Sin∅ = -5/13 , Cosec∅=-13/5 , Tan∅=-5/12 , Cot∅=-12/5 & Sec∅=13/12
Question 39 . SinX=-3/4 X Lies in Quadrant-3(Where only Tan & Cot are Positive all other Functions Are Negative in this Quadrant)
- Now For Other Trignometry Functions
Find By Using Basic Formulas
[tex]SinX^{2}+CosX^{2}=1[/tex] ↔ Put Value of Sin∅=-3/4
We get CosX= ±[tex]\frac{\sqrt{5} }{4}[/tex] (but X Lies in Quadrant-3 Where Cosine Is Negative , So Neglect CosX= +[tex]\frac{\sqrt{5} }{4}[/tex])
So CosX= -[tex]\frac{\sqrt{5} }{4}[/tex] ...... now you get the point . All Questions Will Solve like This .
