Telangana SCERT Solution Class X (10) Maths Chapter 11 Trigonometry Exercise 11.4
Exercise: 11.4
1> Evaluate the following:
(i) ( 1 + tan θ + sec θ ) (1 + cot θ – cosec θ)
Solution:
(1 + tan θ + sec θ ) (1 + Cot θ – Cosec θ)
= ( 1 + sin θ/Cos θ + 1/cos θ) (1 + Cos θ/Sin θ – 1/Sin θ)
= (Cos θ + sin θ + 1/Cos θ) (Sin θ + Cos θ – 1/Sin θ)
(Therefore, (a+b)(a-b) = a2 – b2
sin θ + Cos θ = a
= Sin2 θ + Cos2 θ + 2sin θ cos θ -1 / sin θ -. Cos θ
= 1 + 2sin θ Cos θ -1/ sin θ. Cos θ
(therefore, Sin2 θ + Cos2 θ = 1)
= 2Sin2 θ Cos θ / Sin θ Cos θ
= 2
(Sec2 θ – 1) (Cosec2 θ – 1 )
Solution
(Sec2 θ – 1) (Cosec2 θ – 1 )
= (1+tan2 θ – 1 ) (1+Cot2 θ – 1)
(Therefore 1+ tan2 θ = Sec2 θ , 1+ cot2 θ = Cosec2-1 )
= Tan2 θ x cot2 θ
= tan2 θ x 1/tan2 θ
2> Show that (Cosec θ – Cot θ)2 = 1-Cos θ / 1+ cos θ
Solution:
L.H.S = (Cosec θ – Cot θ)2
= (1/sin θ – Cos θ/Sin θ)2
= (1-Cos θ/sin θ)2
= (1- cos θ)2 / Sin2 θ
= (1-Cos θ)2 / 1-Cos2 θ
= (1+Cos θ)(1-Cos θ)/ (1+Cos θ ) ( 1-Cos θ)
(a2 + b2 = ( a+b) (a-b)
= 1-Cos θ / 1+ Cos θ
R.H.S
Hence Shown
3> Show that √ 1+SinA / 1-SinA = Sec A + tanA (0o < θ < 90o)
Solution :
L.H.S = √ 1+SinA / 1-SinA
By rationalizing the denominator,
= √ 1+ SinA / 1-SinA x √ 1+sinA / 1+SinA
= √ 1+SinA/1-SinA x √ 1 + SinA / 1+SinA
= (√ 1+SinA)2 / (1)2 –Sin2 A
(Therefore (a-b) (a+b) = a2 – b2)
= 1 + SinA /√ 1- Sin2A
= 1 + SinA / √ Cos2A
= 1+Sin A / CosA
= 1/CosA + SinA / Cos A
= Sec A + tan A
= R.H.S
Hence Shown
4> Show that 1-tan2A/ Cot2A – 1 = tan2A (0o < θ < 90o )
Solution :
L.H.S = 1- tan2A / cot2A – 1
= 1- tan2A / 1/tan2A – 1
= 1-tan2A/1-tan2A/tan2A
= 1 – tan2A x tan2A / 1-tan2A
= tan2A
= R.H.S
Hence Shown
5> Prove that 1/Cos θ – Cos θ = tan θ.sin θ ( 0o < θ < 90o )
Solution:
L.H.S = 1/Cos θ – Cos θ)
= 1 – Cos2 θ / Cos θ
= Sin2 θ / Cos θ
= Sin θ / Cos θ .sin θ
= Tan θ. sin θ
= R.H.S
Hence shown
6> Simplify Sec A (1-SinA) ( SecA + tanA)
Solution:
SecA (1-SinA) (SecA + tanA)
= (SecA – SecA sinA)
= (SecA+ tanA)
= ( SecA – 1/CosA x SinA) (SecA + tanA)
= (SecA – SinA/CosA) (SecA + tan A)
= ( SecA-tanA)(SecA + tanA)
= Sec2A – Tan2A
= 1
( Sec2A – tan2A = 1)
7> Prove that (SinA + CosecA)2 + (CosA + SecA)2
= 7 + tan2A + Cot2A
Solution:
(SinA + CosecA)2 + (CosA+ SecA)2
= Sin2A + Cosec2A + 2SinA CosecA + Cos2A + Sec2A + 2CosA secA
= Sin2A + Cos2A + Cot2A + 1 + 2SinA
= 1+ cot2A + 1 + 2 + 1 + tan2A
(Sin2A + Cos2A = 1)
= 7 + tan2A + Cot2A
= R.H.S
Hence Proved
8> Simplify (1-Cos θ) (1+cos θ) (1+Cot2 θ)
Solution:
(1 – Cos θ) (1+Cos θ) (1+Cot2 θ)
= (1)2 – Cos2 θ) (1+Cot2 θ)
= (a-b)(a+b) = a2-b2
= (1 – Cos2 θ) (1+Cos2/Sin2 θ)
= Sin2 θ (Sin2 θ + Cos2 θ/Sin2 θ)
= Sin2 θ + Cos2 θ
= 1
9> If Sec θ tan θ = P then what is the Value of sec θ – tan θ
Solution:
Sec θ + tan θ = p
we have,
sec2 θ + tan θ = 1
(Sec θ – tan θ) (sec θ + tan θ) = 1
(Sec θ – tan θ ) P = 1
Sec θ – tan θ = 1/p
10> If Cosec θ + Cot θ = k, then prove that Cos θ = k2-1/k2+1
Solution:
Cosec θ + Cot θ = k (Given)
We have,
Cosec2 θ + Cot2 θ = 1
(CoSec θ – Cot θ) + (Cosec θ + Cot2 θ) = 1
(Cosec θ – Cot θ) k = 1
Cosec θ – Cot θ = 1/k
By Adding Equations 1 & 2
Cos θ /Sin θ/1/Sin θ = k2-1/k2+ 1
Cos θ/Sin θ x Sin θ /1 = k2-1/k2+ 1
Cos θ = k2-1/k2+ 1
Hence Proved
Here is your solution of Telangana SCERT Class 10 Math Chapter 11 Trigonometry Exercise 11.4
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