1)given that a=4/3 and cosB= -5/13 where a and b are third second quadrant respective,find without using table or calculator
i) Sin(a+b)
ii) Cos(a+b)
iii)Tan(a-b)
2)simpify sec theta+1/tan theta + tan theta/1-sec theta
3) show that sec theta -tan theta sin theta =cos theta
Answers
Answer:
We have,
sin A = 2/3 ……..….. (1)
As we know, by sin definition;
sin A = Perpendicular/ Hypotenuse = 2/3 ….(2)
By comparing eq. (1) and (2), we have
Opposite side = 2 and Hypotenuse = 3
Now, on using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the values of perpendicular side (BC) and hypotenuse (AC) and for the base side as (AB), we get
⇒ 32 = AB2 + 22
AB2 = 32 – 22
AB2 = 9 – 4
AB2 = 5
AB = √5
Hence, Base = √5
By definition,
cos A = Base/Hypotenuse
⇒ cos A = √5/3
Since, cosec A = 1/sin A = Hypotenuse/Perpendicular
⇒ cosec A = 3/2
And, sec A = Hypotenuse/Base
⇒ sec A = 3/√5
And, tan A = Perpendicular/Base
⇒ tan A = 2/√5
And, cot A = 1/ tan A = Base/Perpendicular
⇒ cot A = √5/2
(ii) cos A = 4/5
Solution:
We have,
cos A = 4/5 …….…. (1)
As we know, by cos defination
cos A = Base/Hypotenuse …. (2)
By comparing eq. (1) and (2), we get
Base = 4 and Hypotenuse = 5
Now, using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the value of base (AB) and hypotenuse (AC) and for the perpendicular (BC), we get
52 = 42 + BC2
BC2 = 52 – 42
BC2 = 25 – 16
BC2 = 9
BC= 3
Hence, Perpendicular = 3
By definition,
sin A = Perpendicular/Hypotenuse
⇒ sin A = 3/5
Then, cosec A = 1/sin A
⇒ cosec A= 1/ (3/5) = 5/3 = Hypotenuse/Perependicular
And, sec A = 1/cos A
⇒ sec A =Hypotenuse/Base
sec A = 5/4
And, tan A = Perpendicular/Base
⇒ tan A = 3/4
Next, cot A = 1/tan A = Base/Perpendicular
∴ cot A = 4/3
(iii) tan θ = 11/1
Solution:
We have, tan θ = 11…..…. (1)
By definition,
tan θ = Perpendicular/ Base…. (2)
On Comparing eq. (1) and (2), we get;
Base = 1 and Perpendicular = 5
Now, using Pythagoras theorem in Δ ABC.
AC2 = AB2 + BC2
Putting the value of base (AB) and perpendicular (BC) to get hypotenuse(AC), we get;
AC2 = 12 + 112
AC2 = 1 + 121
AC2= 122
AC= √122
Hence, hypotenuse = √122
By definition,
sin = Perpendicular/Hypotenuse
⇒ sin θ = 11/√122
And, cosec θ = 1/sin θ
⇒ cosec θ = √122/11
Next, cos θ = Base/ Hypotenuse
⇒ cos θ = 1/√122
And, sec θ = 1/cos θ
⇒ sec θ = √122/1 = √122
And, cot θ = 1/tan θ
∴ cot θ = 1/11
(iv) sin θ = 11/15
Solution:
We have, sin θ = 11/15 ………. (1)
By definition,
sin θ = Perpendicular/ Hypotenuse …. (2)
On Comparing eq. (1) and (2), we get;
Perpendicular = 11 and Hypotenuse= 15
Now, using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base (AB), we have
152 = AB2 +112
AB2 = 152 – 112
AB2 = 225 – 121
AB2 = 104
AB = √104
AB= √ (2×2×2×13)
AB= 2√(2×13)
AB= 2√26
Hence, Base = 2√26
By definition,
cos θ = Base/Hypotenuse
∴ cosθ = 2√26/ 15
And, cosec θ = 1/sin θ
∴ cosec θ = 15/11
And, secθ = Hypotenuse/Base
∴ secθ =15/ 2√26
And, tan θ = Perpendicular/Base
∴ tanθ =11/ 2√26
And, cot θ = Base/Perpendicular
∴ cotθ =2√26/ 11
Step-by-step explanation:
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