Telangana SCERT Solution Class X (10) Maths Chapter 11 Trigonometry Exercise 11.3
Exercise 11.3
Evaluate
1> tan36o/Cot54o
Solution : tan36o/Cot54o
= tan (90o – 54o)/Cot54o
= Cot54o/Cot54o
(Therefore, tan(90o – θ) = (0+θ)
Tan36o/Cot54o=1
2> Cos12o – Sin78o
Solution:
Cos12o – Sin78o = Cos(90-78o)
= Sin78o – Sin78o
(Cos (90 – θ) = sinθ
Cos12o – Sin78o = 0
3> Cosec31o – sec59o
Solution:
Cosec31o – Sec59o
= Cosec(90o – 59o) – Sec59o
= Sec59o – Sec59o
(Therefore, Cosec(90o – θ) = Secθ
Cosec31o – Sec59o = 0
4> Sin15o Sec75o
Solution:
Sin15o Sec75o
= Sin (90o-75o) Sec75o
= Cos 75o Sec75o
(Therefore Sin (90o – θ) = Cosθ
= 1/Sec75o x Sec75o
Sin15o Sec75o = 1
5> Tan26o tan54o
Solution:
Tan26o tan54o = tan(90o – 54o) Tan54o
(Therefore tan(90o – θ) = Cotθ
= 1/tan54o x tan54o
tan26o tan54o = 1
2> Show that tan48o tan16o tan42o tan74o = 1
Solution:
L.H.S = Tan48o tan16o tan420 tan74o
= Tan (90o – 42o) tan 42o tan (90o – 74o)
= Cot 42o x tan42o x Cot74o tan74o
(Therefore, Tan (90o – θ) = (0 + θ)
L.H.S = 1/tan42o x tan42o x 1/tan74o x tan74o
= Y
(Therefore, Cotθ = 1/tanθ)
= R.H.S
3> Cos36o Cos54o – Sin36o sin54o = 0
Solution :
L.H.S = Cos36o Cos54o – Sin36o Sin54o
= Cos (90o -54o) Cos 54o – Sin (90o – 54o)
= Sin54o Cos54o – Cos 54o Sin54o
(Therefore, Cos (90-θ) = Sin θ, Sin (90 – θ) = Cosθ
= 0
= R.H.S
Hence Shown
3> If tan 2A = Cot (A-18o) where 2A is an a Cute angle. Find the value of A
Solution:
Tan 2A = Cot (A-18o)
= Cot (90 – 2A)
= Cot (A-18o)
Therefore, Tanθ = Cot (90 – θ)
Removing Cot on Both Sides
90- 2A = A -18o
90 + 18 = A + 2A
3A = 108o
A = 108o / 3
A = 36o
4> If tan A = CotB where A and B are acute Angle, Prove that A + B = 90o
Solution:
TanA = CotB (Therefore, Given)
tan A = tan (90 – θ)
Therefore, Cot θ = tan (90- θ)
Removing ‘tan’ on both sides
A = 90 – B
A + B = 90o
5> If A, B and C are interlar angles of a triangle ABC, then Show that tan (A+b/2) = Cot C/2
Solution :
A + B + C = 180o
A + B = 180o – C
Dividing the interiar equation by 2
A + B / 2 = 180o – C/ 2
A+b/2 = 180/2 – c/2
A+b/2 = 90o – c/2
By applying ‘ tan’ on both side,
tan ( a+b/2) = tan (90 – c/2)
tan (A+B/2) = Co + c/2
(Therefore (90- θ) = (0+ θ)
6> Express Sin 75o + Cos65o in terms of trigonometric ratios of angles between 0o + 45o
Solution:
Sin75o + Cos65o = Sin (90o – 15) + Cos (90 – 25)
= Cos 15o + Sin25o
(Therefore Sin (90- θ) = Cos θ, Cos (90- θ) = Sin θ
Sin75o + cos65o = Cos15o + Sin25o
Cos 15o lies between 0 & 45o
Sin 25o lies between 0 & 45o
Here is your solution of Telangana SCERT Class 10 Maths Solution Chapter 11 Trigonometry Exercise 11.3
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