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1. sin^2⁡θ+cos^2⁡θ=1
2. cosec^2⁡θ-cot^2⁡θ=1
3. sec^2⁡θ=tan^2⁡θ=1
4. sin⁡(90°-θ)=cos⁡θ
5. cos⁡(90°-θ)=sin⁡θ
6. tan⁡(90°-θ)=cot⁡θ⇒ cot⁡〖(90°-θ)=tan⁡θ 〗
7. cosec⁡(90°-θ)=sec⁡θ
8. sec⁡(90°-θ)=cosec⁡θ

Example 1: In a ΔABC right angled at B if AB = 12, and BC = 5 find sin A and tan A, cos C and cot C
Solution:

AC=√((AB)^2+(BC)^2 )
=√(〖12〗^2+5^2 )
=√(144+25)
=√169=13
When we consider t-ratios of∠A we have
Base AB = 12
Perpendicular = BC = 5
Hypotenuse = AC = 13
sin⁡A=Perpendicular/Hypotenuse=5/13
tan⁡A=Perpendicular/Base=5/12
When we consider t-ratios of ∠C, we have
Base = BC = 5
Perpendicular = AB = 12
Hypotenuse = AC = 13
cos⁡C=Base/Hypotenuse=5/13
cot⁡C=Base/Perpendicular=5/12

Example 2: In a right triangle ABC right angle at B the six trigonometric ratios of C
Solution:
sin⁡A=Perpendicular/Hypotenuse=3/5

Base=√((Hypotenuse)^2-(Perpendicualr)^2 )
=√(5^2-3^2 )
=√(25-9)=√16=4
Now
sin⁡C=BC/AC=4/5,cosec⁡C=5/4
cos⁡C=3/5=AB/AC,sec⁡C=5/3
tan⁡C=AB/AC=4/3,cot⁡C=3/4

Example 3: Find the value of 2 sin2 30° tan 60° – 3 cos2 60° sec2 30°
Solution:
2(1/2)^2×√3-3(1/2)^2×(2/√3)^2
=2×1/4×√3-3×1/4×4/3=√3/2-1=(√3-2)/2

Example 4:bFind the value θ sin2θ=√3
Solution:
sin⁡2θ= √3/2
2θ = 60
θ = 30°

Example 5: Find the value of x. Tan 3x = sin 45° cos 45° + sin 30°
Solution:
tan⁡3x=1/√2×1/√2+1/2
=1/2+1/2=1
⇒tan⁡3x=1 ⇒tan⁡3x=tan⁡〖45°〗
3x = 45°
X = 15°