1. In D ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) SinA , CosA
(ii) SinC , CosC.
2. In the figure, find tan P – cot R.
3. sin A =
,calculate cos A and tan A.
4. Given 15 cot A = 8, find sin A and sec A.
5. Given secq = 
calculate all other trigonometric Ratios.
6. If Ð A and Ð B are acute angles such that cos A = cos B,then show that Ð A = Ð B.
7. If cot q = 
,evaluate : (i) 
(ii) cot2q.
8. If 3 cotA = 4, Check whether 
=cos2A – sin2A or not.
9. In triangle ABC, right-angled at B, if tan A =
find the value of (i) sin A cos C +cos A sin C (ii)cos A cos C – sin A sin C.
10. In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
(i) SinA , CosA
(ii) SinC , CosC.
- Solution:
2. In the figure, find tan P – cot R.- Solution:In a right angled D PQR
PR2 = PQ2 + QR2132 = 122 + QR2QR2 = 169 – 144 = 25
QR = 5
\tan P - cot R =
= 0
3. sin A =,calculate cos A and tan A.- Solution:Let the right angled D ABC be right angled at B.
In a right angled D ABC
AC2 = AB2 + BC216 = AB2 + BC
AB2 = 16 – 9 = 7
AB =
sin A =
cos A =
= 
tan A =
.
4. Given 15 cot A = 8, find sin A and sec A.- Solution:15 cot A = 8
Þ cot A =
Let ABC be the D right angled at B
Cot A = 8/15
Þ AB = 8, BC = 15
AC2 = AB2 + BC2
= 82 + 152
= 64 + 225
= 289
AC =
= 17
sin A =
= 
sec A =
= 
5. Given secq = calculate all other trigonometric Ratios.- Solution:
In a right angled D ABC,
AC2 = AB2 + BC2132 - 122 = AB2AB2 = 169 - 144
AB = 5
6. If Ð A and Ð B are acute angles such that cos A = cos B,then show that Ð A = Ð B.- Solution:If cosA = cosB
Þcos A – cos B = 0
only when
= 450.
7. If cot q = ,evaluate : (i) (ii) cot2q.- Solution:If cot q =
(i) = 
= 
(ii) cot2q =
.
8. If 3 cotA = 4, Check whether =cos2A – sin2A or not.- Solution:If 3 cot A = 4
= 
= 
= 
………………… (1)
cos2A – sin2A = sin2A(cot2A - 1) Dividing and multiplying by sin2A
= sin2A
= sin2A
= sin2A
=
= 
…………………………. (2)
Since (1) = (2)
LHS = RHS.
9. In triangle ABC, right-angled at B, if tan A =find the value of (i) sin A cos C +cos A sin C (ii)cos A cos C – sin A sin C.- Solution:If tan A =
In a right angled D ABC, right angled at B
AC2 = AB2 + BC2
\AC =
= 2
sin A =
, cos A = 
sin C = 
cos C = 1/2
(i) sin A cos C + cos A sin C
=
= 
= 1
(ii) cos A cos C – sin A sin C
= 0.
10. In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.- Solution:
PR + QR = 25
PQ = 5 cm
In a right angled D PQR, PQ2 + QR2 = PR2 …………………… (1)
Let PR = x
QR = (25 – x)
From 1 x2 = 52 + (25 – x)2
x2 = 25 + (25)2 – 2 ´ 25x + x2
= 625 + 25 – 50x
50x = 650
x =
= 13
PR = 13 cm
QR = 12 cm
PQ = 5 cm
sin P =
= 
cos P =
= 
- tan P =
= 
. - 11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A =
for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin q =
.- Solution:(i) False. The value of tan 90◦ is grater than 1
(ii) True. sec A =
Þ cos A = 
as 12 the hypotenuse is the largest side of triangle.
(iii) False. cos A is the abbreviation for cosine of angle A
(iv) No,False. cos A is not the product of cot and A
(v) False, Because the hypotenuse is the longest side whereas in sin q =
, 3 which is the denominator is cannot be the hypothenuse.
12. Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°
(iii)
(iv)
(v)
.- Solution:
=
13. Choose the correct option and justify your choice:(i) 
=(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii)
=(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv)
=(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°.- Solution:(i)
= 
= 
=
=
= 
= sin 600
Answer: A
(ii)
= 
= 0
Answer: D
(iii) sin 2A = 2 sin A is true when A = 450Answer: A
(iv)
= 
= 
=
=
=
=
´ 
=
= 
= tan 600.Answer: C
14. tan (A + B) = 
and tan (A – B) =
; 0° < A + B 
90°; A > B, find A and B.- Solution:If tan(A + B) =
tan(A - B) = 1/
If tan A+B =
Þ A + B = 600 ………………………(1)
tan (A – B) =
A – B = 300 ……………………..(2)
Adding (1) ´ (2) 2A = 900A = 450
\B = 150.
15. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin
increases as increases.
(iii) The value of cos increases as increases.
(iv) sin = cos for all values of .
(v) cot A is not defined for A = 0°.- Solution:(i) sin(A+B) = sin A + sin B
False
(ii) Value of sin q as q increase
True
(iii) Value of cos q as q ncreases
False
(iv) sinq = cosq for all values of q
False
(v) cot A is not defined for A = 00
True.
16. Evaluate:
(i)
(ii)
(iii) cos480 – sin 420
(iv)cosec 310 – sec 590.- Solution:(i)
=
=
= 1
(ii)
= 
= 
= 1
(iii) cos480 – sin 420cos480 = cos(90 - 420)
= sin 420
\cos 480 – sin 420 = sin 420 – sin 420 = 0
(iv) cosec 310 – sec 590
cosec 310 = cosec(900 - 590)
= sec 590
LHS = sec 590 – sec 590 = 0.
21. A, B and C are interior angles of a triangle ABC, then show that sin
= cos 
.- Solution:
22. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.- Solution:
23. Express the trigonometric ratios sin A, cos A and tan A in terms of sec A.- Solution:
24. Write all the other trigonometric ratios of Ð A in terms of sec A.- Solution:sin A =
=
cos A =
tan A =
= 
¸ 
=
´ sec A.=
25. Evaluate:(i)
(ii) sin 250. cos 650 + cos 250 sin 650.- Solution:
26. Choose the correct option. Justify your choice.
(i) 9 sec2 A – 9 tan2 A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan q + secq ) (1 + cot q – cosec q ) =
(A) 0 (B) 1 (C) 2 (D) –1
(iii) (sec A + tan A) (1 – sin A) =
(A)sec A (B) sin A (C) cosec A (D) cos A
(iv)
(A) sec2 A (B) –1 (C) cot2 A (D) tan2 A.- Solution:(i) 9 sec2A - 9 tan2A
= 9 (1 + tan2A) – 9 tan2A
= 9 + 9 tan2A - 9 tan2A
= 9Answer (B)(ii)
(iii) (sec A + tan A) (1 – sin A)
sec A – sin A sec A + tan A – tan A sin A
+ 
- 
.sinA 
(iv)
= 
= 
= tan2A.
27. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosecq - cot q )2 =
(ii)
= 2 secA
(iii)
= 1 +secq cosecq
(iv)
=
(v)
= Cosec A +cot A- Solution:(i) (cosecq - cot q )2 =
RHS = 
=
= cosec2q - 2cosecq . cotq + cot2q = (cosec q - cotq )2
Hence Proved.(ii)
=
= 
=
= 
=
= 2 sec A
(iii)
(iv)
= 
LHS
RHS
=
= LHS
(v)
Using the identity cosec2A = 1 + cot2A.
Taking the LHS of the given equation
Dividing the numerator and denominator by sin A
=
(Changing 
= cot A ´ 
=
(Replacing cosec2A - cot2A = (cosec A + cot A) (cosec A - cot A) using the identity cosec2A - cot2A = 1
=
=
= cosec A + cotA.
28. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i)
= secA + tanA
- Solution:(i)
=
Taking the conjugate of (1-sinA) + multiplying the numerator and denominator by it.
=
=
(Since 1 - sin2A = cos2A)
=
= 
=
= sec A + tan A (As 
= sec A, 
= tan A)
(ii) (sin A + cosec A)2 + (cos A + sec A)2
= sin2A + cosec2A + 2sinA.cosec A + cos2A + sec2A + 2cos A. sec A
= sin2A + cos2A + cosec2A + sec2A + 2sinA .
+ 2cos A. 
= 1 + 2 + 2 + cosec2A + sec2A (using cosec A =
secA = 
)
= 5 + 1 + tan2A + 1 + cot2A
= 7 + tan2A + cot2A (1 + tan2A = sec2A) 1 + cot2A = cosec2A
(iii) (cosec A - sin A) (secA - cosA)
LHS
(cosec A - sin A) (sec A - cos A)
(since 1 - sin2A = cos2A and 1 - cos2A = sin2A)
=
= sin A cosA
RHS
= 
=
(Taking LCM as cosA. sinA)
=
(since sin2A + cos2A = 1)
LHS = RHS
Hence the result.
(iv)
LHS
RHS
Thus, LHS = RHS.
17. Show that tan 480 tan 230 tan 420 tan 670 = 1.- Solution:(i) tan 480 tan230 tan 420 tan670 = 1
LHS = tan 480. tan 420 tan 230 tan670
= tan 480. tan(900 - 480). tan(900 - 670) tan 670
= tan 480. cot 480. cot670 tan 670 .
= 1 ´ 1 = 1.
18. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.- Solution:tan 2A = cot (A – 18) ………………. (1)
where 2A is an acute angle
tan 2A = cot(90 – 2A) = cot(A – 18)
cot 90 – 2A and A - 18 are acute angles.
90 – 2A = A – 18
3A = 108
A = 360.
19. If tan A = cot B, prove that A + B = 90°.- Solution:If tan A = cot B …………… (1)
tan A = cot (90 – A)
From (1)
cot (90 - A) = cot B
Þ 90 – A = B
or A + B = 900.
20. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.- Solution:If sec 4A = cosec (A – 200)
sec 4A = cosec (90 – 4A)
\cosec(90 – 4A) = cosec (A – 200)
Since (90 – 4A) and A – 200 are acute angles
90 – 4A = A – 200
90 + 20 = 5A
A =
= 220.
- Solution:(i) False. The value of tan 90◦ is grater than 1
- tan P =
= 
.