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Page No 554:

Question 27:

If 3x=cosecθ and 3x=cotθ, find the value of 3x2-1x2.             [CBSE 2010]

Answer:

3x2-1x2=93x2-1x2=139x2-9x2=133x2-3x2
=13cosecθ2-cotθ2=13cosec2θ-cot2θ=131=13

Page No 554:

Question 28:

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find (i) BC, (ii) AB.

Answer:

From the given right-angled triangle, we have:
BCAC = sin 30oBC20 = 12 BC = 202 = 10 cm Also, ABAC = cos 30oAB20 = 32 AB = 20×32 = 103  cm BC = 10 cm and AB = 103  cm

Page No 554:

Question 29:

In the adjoining figure, ∆ABC is right-angled at B and ∠A = 30°. If BC = 6 cm, find (i) AB, (ii) AC.

Answer:

From the given right-angled triangle, we have:

BCAB=tan 30o6AB=13  AB=63 cmAlso, BCAC=sin 30o6AC=12  AC=2×6=12 cmAB = 63 cm and AC = 12 cm

Page No 554:

Question 30:

In the adjoining figure, ∆ABC is a right-angled at B and ∠A = 45°. If AC32 cm,
find (i) BC, (ii) AB.

Answer:

From right-angled ∆ABC, we have:

    BCAC=sin 45oBC32=12  BC=3 cm  Also, ABAC=cos 45oAB32=12  AB=3 cm BC=3 cm and AB=3 cm



Page No 572:

Question 1:

sin 60° cos 30° + cos 60° sin 30°

Answer:

On substituting the values of various T-ratios, we get:
 sin 60o cos 30o + cos 60o sin 30o 
 =32×32+12×12=34+14=44=1

Page No 572:

Question 2:

sin 60° cos 30° − cos 60° sin 30°

Answer:

Ans

Page No 572:

Question 3:

cos 60° cos 30° − sin 60° sin 30°

Answer:

On substituting the values of various T-ratios, we get:
 cos 60o cos 30o − sin 60o sin 30o 
 =12×32-32×12=34-34=0

Page No 572:

Question 4:

cos 45° cos 30° + sin 45° sin 30°

Answer:

On substituting the values of various T-ratios, we get:
 cos 45o cos 30o + sin 45o  sin 30o 
 = 12×32 + 12×12 = 322 + 122 =3 +122

Page No 572:

Question 5:

tan 30° cosec 60° + tan 60° sec 30°

Answer:

As we know that,
tan30°=13sec30°=23cosec60°=23tan60°=3By substituting these values, we gettan30° cosec60°+tan60° sec30°=13×23+3×23                                                =23+2                                                =2+63                                                =83Hence, tan30° cosec60°+tan60° sec30°=83
 

Page No 572:

Question 6:

2 cos2 60° + 3 sin2 45° − 3 sin2 30° + 2 cos2 90°

Answer:

On substituting the values of various T-ratios, we get:
 2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o
 =2×122+3×122-3×122 + 2×02=2×14+3×12-3×14+0= 12+32-34=2+6-34=54

Page No 572:

Question 7:

sin230° cos245°+4tan230°+12sin290°+18cot260°

Answer:

As we know that,
sin30°=12cos45°=12tan30°=13sin90°=1cot60°=13By substituting these values, we getsin230° cos245°+4tan230°+12sin290°+18cot260°=122×122+4132+1212+18132                                                                          =14×12+413+121+1813                                                                          =18+43+12+124                                                                          =3+32+12+124                                                                          =4824                                                                          =2Hence, sin230° cos245°+4tan230°+12sin290°+18cot260°=2.

Page No 572:

Question 8:

cot230° − 2cos230° − 34 sec245° + 14 cosec230°

Answer:

On substituting the values of various T-ratios, we get:
  cot2 30o − 2 cos2 30o − 34 sec2 45o + 14 cosec2 30o
 = 32-2×322-34×22+14×22= 3-2×34-34×2+14×4= 3-32-32+1 = 4-32+32 = 4-3= 1

Page No 572:

Question 9:

(cos 0° + cos 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

Answer:

As we know that,
cos0°=1cos45°=12sin30°=12sin90°=1cos60°=12By substituting these values, we getcos0°+cos45°+sin30°sin90°-cos45°+cos60°=1+12+121-12+12                                                                             =1+12+121+12-12                                                                             =1+122-122                                                                             =1+14+2112-12                                                                             =1+14+1-12                                                                             =4+1+4-24                                                                             =74Hence, cos0°+cos45°+sin30°sin90°-cos45°+cos60°=74.

Page No 572:

Question 10:

Show that:
(i) 1-sin60°cos60°=tan60°-1tan60°+1
(ii) cos30°+sin60°1+sin30°+cos60°=cos30°

Answer:

(i)
 LHS=1-sin 60ocos 60o=1-3212=2-3212=2-32×2=2-3RHS= tan 60o-1tan 60o+1=3-13+1=3-13+1×3 -13 -1=3-1232-12=3+1-233-1=4-232=2-3

Hence, LHS = RHS

 1-sin 60ocos 60o=tan 60o-1tan 60o+1

(ii)
 LHS = cos 30o+sin 60o1+sin 30o+cos 60o=32+32 1+12+12 =3+322+1+12=32Also, RHS = cos 30o=32

Hence, LHS = RHS

   cos 30o+sin 60o1+sin 30o+cos 60o=cos 30o1sin60°cos60°

Page No 572:

Question 11:

Verify each of the following:
(i) sin 60° cos 30° − cos 60° sin 30° = sin 30°
(ii) cos 60° cos 30° + sin 60° sin 30° = cos 30°
(iii) 2 sin 30° cos 30° = sin 60°
(iv) 2 sin 45° cos 45° = sin 90°

Answer:

(i) sin 60o cos 30o − cos 60o sin 30o
 =32×32-12×12=34-14=24=12Also, sin 30o=12
∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o

=12×32+32×12=34+34=32Also, cos 30o =32
∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o
=2×12×32=32Also, sin 60o =32
∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
=2×12×12=1
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o

Page No 572:

Question 12:

If A = 45°, verify that:
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = 2 cos2 A − 1 = 1 − 2 sin2 A

Answer:

A = 45o
⇒ 2A = 2 × 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o 2×12×12 = 1
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2 A − 1 = 2 cos2 45o − 1 =  2×122 - 1 = 2×12 -1 = 1-1 = 0
Now, 1 − 2 sin2 A1-2×122 = 1 - 2×12  =1 - 1 = 0
∴ cos 2A 2 cos2 A − 1 = 1 − 2 sin2 A

Page No 572:

Question 13:

If A = 30°, verify that:
(i) sin2A=2tanA1+tan2A
(ii) cos2A=1-tan2A1+tan2A
(iii) tan2A=2tanA1-tan2A

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

​(i) sin 2A = sin 60o = 32
2 tan A1+tan2 A=2 tan 30o1+tan2 30o=2×131+132 =231 + 13=2343=23×34=32
sin 2A=2tan A1+tan2A

(ii) cos 2A = cos 60o = 12

1-tan2 A1+tan2 A=1-tan2 30o1+tan2 30o=1-1321+132 =1 - 131 + 13=2343=23×34=12
cos 2A=1-tan2 A1+tan2 A

(iii) tan 2A = tan 60o = 3
2 tan A1-tan2 A=2 tan 30o1-tan2 30o=2×131-132 =231-13=2323=23×32=3
tan 2A=2 tan A1-tan2 A=2tanA1+tan2A

Page No 572:

Question 14:

If A = 60° and B = 30°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B − sin A sin B

Answer:

A = 60o and B = 30o
Now, A + B = 60o + 30o​ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o
32×32+12×12=34 + 14=1
∴ sin (A + B) = sin A cos B + cos A sin B


(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o  =12×32 - 32×12 = 34 - 34 = 0
∴​ cos (A + B) = cos A cos B − sin A sin B
 



Page No 573:

Question 15:

If A = 60° and B = 30°, verify that:

(i) sin (A B) = sin A cos B − cos A sin B
(ii) cos (AB) = cos A cos B + sin A sin B
(iii) tan (A − B) = tan A-tan B1+tan A tan B

Answer:

(i) sin (A − B) = sin 30o = 12
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o
32× 32- 12×12 = 34 - 14 = 24 = 12
∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (AB) = cos 30o = 32
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
= 12×32 + 32×12 = 34 + 34 = 2×34 = 32
∴​ cos (AB) = cos A cos B + sin A sin B

(iii) tan (AB) = tan 30o = 13
 tan A-tan B1+tan A tan B=tan 60o- tan 30o1 +tan 60o tan 30o= 3 -13 1+3×13=12×3- 13=13
∴​ tan (AB) = tan A-tan B1+tan A tan B

Page No 573:

Question 16:

If A and B are acute angles such that tan A = 13, tan B12 and tan (A + B) = tan A + tan B1-tan A tan B , show that A + B = 45°.

Answer:

Given:
tan A = 13 and tan B= 12tan (A+B) = tan A+tan B1-tan A tan BOn substituting these values in RHS of the expression, we get:tan A+tan B1-tan A tan B=13+121-13×12=561-16=5656=1 tan (A+B)=1=tan 45o       [ tan 45o =1]A+B=45o

tan A = 13 and tan B = 12By substituting the values, we get;tan (A+ B) = tan A + tan B1- tan A tan B = 13 +121-13×12 = 561-16 = 5656 = 1tan (A + B) = 1 = tan 45oHence, A + B = 4513

Page No 573:

Question 17:

Using the formula, tan 2 A=2tanA1-tan2A , find the value of tan 60°, it being given that tan 30° = 13.

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:
tan 2A = 2 tan A1-tan2 A tan 60o=2 tan 30o1-tan2 30o=2×131-132=231-13=2323=23×32=3
∴ tan 60o = 3

Page No 573:

Question 18:

Using the formula, cos A=1+cos2A2 , find the value of cos 30°, it being given that cos 60° = 12.

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:
cos A=1+cos 2A2 cos 30o=1+ cos 60o2=1+122=322  =34=32
∴ cos 30o = 32

Page No 573:

Question 19:

Using the formula, sin A=1-cos2A2 , find the value of sin 30°, it being given that cos 60° = 12.

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:

sin A=1-cos 2A2sin 30o=1-cos 60o2=1-122=122  =14=12
∴ sin 30o = 12

Page No 573:

Question 20:

If sinA+B=sinA cosB+cosA sinB and cosA-B=cosA cosB+sinA sinB, find the values of i sin75° and ii cos15°.

Answer:

Let A=45° and B=30°

iAs, sinA+B=sinA cosB+cosA sinBsin45°+30°=sin45° cos30°+cos45° sin30°sin75°=12×32+12×12sin75°=322+122 sin75°=3+122

iiAs, cosA-B=cosA cosB+sinA sinBcos45°-30°=cos45° cos30°+sin45° sin30°cos15°=12×32+12×12cos15°=322+122 cos15°=3+122

Disclaimer: cos15° can also be calculated by taking A=60° and B=45°.

Page No 573:

Question 21:

If tan (x + 30°) = 1 then find the value of x.

Answer:

As we know that,
tan45°=1Thus,if tanx+30°=1x+30°=45°x=45°-30°x=15°Hence, the value of x is 15°.

Page No 573:

Question 22:

If cot θ-1cot θ+1=1-31+3 then find the acute angle θ.

Answer:

Given: cotθ-1cotθ+1=1-31+3cotθ-1cotθ+1=1-31+31tanθ-11tanθ+1=1-31+3              cotθ=1tanθ1-tanθtanθ1+tanθtanθ=1-31+31-tanθ1+tanθ=1-31+3On comparing LHS and RHS, we gettanθ=3θ=60°            tan60°=3Hence, the acute angle θ is equal to 60°.

Page No 573:

Question 23:

If sin (A + B) = 1 and tan A-B=13, 0°<A+B90° and A>B then find the values of A and B.

Answer:

As we know that,
sin90°=1Thus,if sinA+B=1A+B=90°          ...1and tan30°=13Thus,if tanA-B=13A-B=30°          ...2Solving 1 and 2, we getA=60° and B=30°Hence, the values of A and B are 60° and 30°,respectively.

Page No 573:

Question 24:

If sinA+B=32 and cosA-B=32, 0°<A+B90° and A>B then find the values of A and B.

Answer:

As we know that,
sin60°=32Thus,if sinA+B=32A+B=60°          ...1and cos30°=32Thus,if cosA-B=32A-B=30°          ...2Solving 1 and 2, we getA=45° and B=15°Hence, the values of A and B are 45° and 15°,respectively.

Page No 573:

Question 25:

If tan (A − B) = 13 and tan (A + B) = 3, 0° < (A + B) < 90° and A > B, then find A and B.

Answer:

Here, tan (A − B) = 13 
⇒ tan (A B) = tan 30o       [∵ tan 30o = 13]
A − B = 30o                     ...(i)

Also, tan (A + B) = 3 
⇒​ tan (A + B) =  tan 60o        [∵ tan 60o = 3]
A + B = 60o                           ...(ii)  

Solving (i) and (ii), we get:
A = 45o and B = 15o



Page No 574:

Question 26:

If cosec (A + B) = 1 and cosec(A – B) = 2, 0° < (A + B) ≤ 90° and A > B then find the values of :
(i) sin A cos B + cos A sin B
(ii) tan A-tan B1+tan A tan B

Answer:

As we know that,
cosec90°=1Thus,if cosecA+B=1A+B=90°          ...1and cosec30°=2Thus,if cosecA-B=2A-B=30°          ...2Solving 1 and 2, we getA=60° and B=30°           ...3Now,(i) sin60°=32cos30°=32cos60°=12sin30°=12On substituting these values, we getsinA cosB+cosA sinB=sin60° cos30°+cos60° sin30°                                 =32×32+12×12                                 =34+14                                 =44                                 =1Hence, sinA cosB+cosA sinB=1.(ii) tan60°=3tan30°=13On substituting these values, we gettanA-tanB1+tanA tanB=tan60°-tan30°1+tan60° tan30°                      =3-131+313                      =3-131+1                      =223                      =13Hence, tanA-tanB1+tanA tanB=13.

Page No 574:

Question 31:

Find the value of x for which
(i) x tan 45° cot 60° = sin 30° cosec 60°
(ii) 2cosec230°+x sin260°-34tan230°=10

Answer:

(i) As we know that,
tan45°=1cot60°=13sin30°=12cosec60°=23On substituting these values, we getxtan45° cot60°=sin30° cosec60°x113=1223x3=13x=1Hence, the value of x is 1.


(ii) As we know that,
tan30°=13sin60°=32cosec30°=2On substituting these values, we get2cosec230°+xsin260°-34tan230°=10222+x322-34132=1024+x34-3413=108+3x4-14=103x4-14=10-83x-14=23x-1=83x=9x=3Hence, the value of x is 3.



Page No 576:

Question 1:

(sec260°– 1) = ?
(a) 0
(b) 2
(c) 3
(d) 4

Answer:

As we know that,
sec60°=2By substituting the value, we getsec260°-1=22-1                   =4-1                   =3Hence, the correct option is c.

Page No 576:

Question 2:

(sin230° – sec260° + 4cot245°) = ?
(a) 4
(b) 2
(c) 1
(d) 14

Answer:

As we know that,
sin30°=12sec60°=2cot45°=1By substituting these values, we getsin230°-sec260°+4cot245°=122-22+412                                            =14-4+4                                            =14Hence, the correct option is d.

Page No 576:

Question 3:

(3cos260° + 2cot230° – 5sin245°) = ?
(a) 1

(b) 4

(c) 174

(d) 136

Answer:

As we know that,
sin45°=12cos60°=12cot30°=3By substituting these values, we get3cos260°+2cot230°-5sin245°=3122+232-5122                                                =314+23-512                                                =34+6-52                                                =3+24-104                                                =174Hence, the correct option is c.

Page No 576:

Question 4:

cos230° cos245°+4sec260°+12cos290°-2tan260°=?

(a) 818

(b) 838

(c) 738

(d) 758

Answer:

As we know that,
cos30°=32cos45°=12sec60°=2cos90°=0tan60°=3By substituting these values, we getcos230° cos245°+4sec260°+12cos290°-2tan260°=322122+422+1202-232                                                                              =3412+44-23                                                                              =38+16-6                                                                              =38+10                                                                              =3+808                                                                              =838Hence, the correct option is b.

Page No 576:

Question 5:

(cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° – cos 45°) = ?

(a) 74

(b) 56

(c) 35

(d) 58

Answer:

As we know that,
cos0°=1cos45°=12sin30°=12sin90°=1cos60°=12sin45°=12By substituting these values, we getcos0°+sin30°+sin45°sin90°+cos60°-cos45°=1+12+121+12-12                                                                            =1+12+121+12-12                                                                            =1+122-122                                                                            =1+14+2112-12                                                                            =1+14+1-12                                                                            =4+1+4-24                                                                            =74Hence, the correct option is a.

Page No 576:

Question 6:

If tan2 45° – cos230° = x sin 45° cos 45° then x = ?

(a) 2

(b) –2

(c) 12

(d) -12

Answer:

As we know that,
cos45°=12cos30°=32tan45°=1sin45°=12By substituting these values, we gettan245°-cos230°=x sin45° cos45°12-322=x12121-34=x24-34=x2x2=14x=12Hence, the correct option is c.

Page No 576:

Question 7:

If 2 sin (60° – α) = 1 then α = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Answer:

As we know that,
sin45°=12Thus,if 2 sin60°-α=1sin60°-α=1260°-α=45°α=60°-45°α=15°Hence, the correct option is a.

Page No 576:

Question 8:

If tan x = 3 cot x then x = ?
(a) 60°
(b) 45°
(c) 30°
(d) 15°

Answer:

Given: tan= 3cotx


tanx=3cotxtanx=31tanx          cotx=1tanxtan2x=3tanx=3x=60°                   tan60°=3  Hence, the correct option is a.

Page No 576:

Question 9:

If 3 tan 2θ-3=0 then θ=?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Answer:

As we know that,
tan60°=3Thus,if 3 tan2θ-3=03 tan2θ=3tan2θ=33tan2θ=32θ=60°θ=30°Hence, the correct option is b.

Page No 576:

Question 10:

If 2sin 2θ=3 then θ=?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

As we know that,
sin60°=32Thus,if 2sin2θ=3sin2θ=322θ=60°θ=30°Hence, the correct option is a.

Page No 576:

Question 11:

If 2cos 3θ=1 then θ=?
(a) 10°
(b) 15°
(c) 20°
(d) 30°

Answer:

As we know that,
cos60°=32Thus,if 2cos3θ=1cos3θ=123θ=60°θ=20°Hence, the correct option is c.

Page No 576:

Question 12:

If x tan 45° cos 60° = sin 60°cot 60° then x = ?
(a) 1

(b) 12

(c) 12

(d) 3

Answer:

As we know that,
tan45°=1cos60°=12sin60°=32cot60°=13By substituting these values, we getx tan45° cos60°=sin60° cot60°x112=3213x2=12x=1Hence, the correct option is a.



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Question 13:

If tan (3x + 30°) = 1 then x = ?
(a) 20
(b) 15°
(c) 10°
(d) 5°

Answer:

As we know that,
tan45°=1Thus,if tan3x+30°=13x+30°=45°3x=45°-30°3x=15°x=5°Hence, the correct option is d.

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Question 14:

If sin θ = cos θ, 0 ≤ θ ≤ 90° then θ = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

Given: sinθ=cosθsinθ=sin90°-θ                 sin90°-θ=cosθθ=90°-θ2θ=90°θ=45°Hence, the correct option is b.

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Question 15:

If sinA-B=12 and cos A+B=12, 0°<A+B90° and A>B then values of A and B are
(a) (60°, 30°)
(b) (60°, 15°)
(c) (45°, 15°)
(d) (60°, 25°)

Answer:

As we know that,
sin30°=12Thus,if sinA-B=12A-B=30°          ...1and cos60°=12Thus,if cosA+B=12A+B=60°          ...2Solving 1 and 2, we getA=45° and B=15°Hence, the correct optiom is c.



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