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(i)
(ii)
,
3. If sin A =3/4, calculate cos A and tan A.
B =90
We know that sin A = BC/AC = 3/4
Let BC be 3k and AC will be 4k where k is a positive real number.
By Pythagoras theorem we get,
AC
2
= AB
2
+ BC
2
(4k)
2
= AB
2
+ (3k)
2
16k
2
- 9k
2
= AB
2
AB
2
= 7k
2
AB = √7 k
cos A = AB/AC = √7 k/4k = √7/4
tan A = BC/AB = 3k/√7 k = 3/√7
4. Given 15 cot A = 8, find sin A and sec A.
Let ΔABC be a right-angled triangle, right-angled at B.
We know that cot A = AB/BC = 8/15 (Given)
Let AB be 8k and BC will be 15k where k is a positive real number.
By Pythagoras theorem we get,
AC
2
= AB
2
+ BC
2
AC
2
= (8k)
2
+ (15k)
2
AC
2
= 64k
2
+ 225k
2
AC
2
= 289k
2
AC = 17 k
sin A = BC/AC = 15k/17k = 15/17
sec A = AC/AB = 17k/8 k = 17/8
5. Given sec θ = 13/12, calculate all other trigonometric ratios.
Let AB = 12k and AC = 13k
Then, using Pythagoras theorem,
(AC)2 = (AB)2 + (BC)2
(13
k
)2 = (12
k
)2 + (BC)2
169
k
2 = 144
k
2 + BC2
25
k
2 = BC2
BC = 5
k
(I) Sin
=
=
=
(ii) Cos θ =
=
=
(iii)Tan θ =
=
=
(iv) Cot θ =
=
=
(v) Cosec θ =
=
=
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
7. If cot θ =7/8, evaluate :
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot
2
θ
(ii)
8. If 3cot A = 4/3 , check whether (1-tan
2
A)/(1+tan
2
A) = cos
2
A – sin
2
A or not.
.
And
9. In triangle ABC, right-angled at B, if tan A =1/√3 find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
.
(i)
(ii)
(ii) 2 tan
2
45° + cos
2
30° – sin
2
60°
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos
2
60° + 4sec
2
30° - tan
2
45°)/(sin
2
30° + cos
2
30°)
2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan
2
30° =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan
2
45°/1+tan
2
45° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin
2
A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan
2
30° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
(ii) 1-tan
2
45°/1+tan
2
45° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin
2
A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan
2
30° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.
……….(i)
4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
2. Write all the other trigonometric ratios of ∠A in terms of sec A.
3. Evaluate :
(i) (sin
2
63° + sin
2
27°)/(cos
2
17° + cos
2
73°)
(ii) sin 25° cos 65° + cos 25° sin 65°
(ii) sin 25° cos 65° + cos 25° sin 65°
4. (i) 9 sec
2
A - 9 tan
2
A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ - cosec θ)
(A) 0 (B) 1 (C) 2 (D) - 1
(iii) (secA + tanA) (1 - sinA) =
(A) secA (B) sinA (C) cosecA (D) cosA
(iv) 1+tan
2
A/1+cot
2
A =
(A) sec
2
A (B) -1 (C) cot
2
A (D) tan
2
A
(ii) (1 + tan θ + sec θ) (1 + cot θ - cosec θ)
(A) 0 (B) 1 (C) 2 (D) - 1
(iii) (secA + tanA) (1 - sinA) =
(A) secA (B) sinA (C) cosecA (D) cosA
(iv) 1+tan
2
A/1+cot
2
A =
(A) sec
2
A (B) -1 (C) cot
2
A (D) tan
2
A
5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
(i) (cosec θ - cot θ)
2
= (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin
2
A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec
2
A = 1+cot
2
A.
(vi) √1 + sin A/1 - sin A = sec A+ tan A
(vii) (sin θ - 2sin
3
θ)/(2cos
3
θ-cos θ) = tan θ
(viii) (sin A + cosec A)
2
+ (cos A + sec A)
2
= 7+tan
2
A+cot
2
A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan
2
A/1+cot
2
A) = (1-tan A/1-cot A)
2
= tan
2
A
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin
2
A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec
2
A = 1+cot
2
A
(vi) √1 + sin A/1 - sin A = sec A+ tan A
LHS = 1 + sin A/(1 - sin A) .....(1)
Multiplying and dividing by (1 + sin A)
⇒ (1 + sin A)(1 + sin A/1 - sin A)(1 + sin A)
= (1 + sin A)²/(1 - sin² A) [a² - b² = (a - b)(a + b)]
= (1 + sinA)/1 - sin² A
= 1 + sin A/cos² A
= 1 + sin A/cos A
= 1/cos A + sin A/cos A
= sec A + tan A
= R.H.S
(vii) (sin θ - 2sin
3
θ)/(2cos
3
θ-cos θ) = tan θ
(viii) (sin A + cosec A)
2
+ (cos A + sec A)
2
= 7+tan
2
A+cot
2
A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan
2
A/1+cot
2
A) = (1-tan A/1-cot A)
2
= tan
2
A
