Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 6 T Ratios Of Some Particular Angles are provided here with simple step-by-step explanations. These solutions for T Ratios Of Some Particular Angles are extremely popular among Class 10 students for Math T Ratios Of Some Particular Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 301:

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 301:

Question 2:

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 301:

Question 3:

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 301:

Question 4:

Evaluate:

sin30°cos45°+cot45°sec60°-sin60°tan45°+cos30°sin90°

Answer:

sin30°cos45°+cot45°sec60°-sin60°tan45°+cos30°sin90°=1212+12-321+321=22+12-32+32=2+12



Page No 302:

Question 5:

Evaluate:

5cos260°+4sec230°-tan245°sin230°+cos230°

Answer:

5cos260°+4sec230°-tan245°sin230°+cos230°=5122+4232-12122+322=54+4×43-114+34=54+163-1144=15+64-121244=67121=6712

Page No 302:

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 302:

Question 7:

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 302:

Question 8:

(sin230° + 4cot245° − sec260°)(cosec245° sec230°)

Answer:

On substituting the values of various T-ratios, we get:
  (sin2 30o + 4 cot2 45o − sec2 60o )(cosec2 45o sec2 30o)
 =122+4×12-22 22 232=14+4-4 2×43 =14×83=23

Page No 302:

Question 9:

4cot230°+1sin230°-2cos245°-sin20°

Answer:

On substituting the values of various T-ratios, we get:
 
 4cot2 30o+1sin2 30o-2 cos2 45o-sin2 0o=432+1122-2×122-02=43+114-2×12-0=43+4-1 =43+3=4+93=133 

Page No 302:

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 302:

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 302:

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 302:

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 302:

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 302:

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 302:

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 302:

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 302:

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 303:

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 303:

Question 20:

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.
Figure

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 303:

Question 21:

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

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 303:

Question 22:

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

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 303:

Question 23:

If sin (A + B) = 1 and cos (A − B) = 1, 0° ≤ (A + B) ≤ 90° and A > B, then find A and B.

Answer:

Here, sin (A + B) = 1
⇒ sin (A+ B) = sin 90o                     [∵ sin 90o = 1]
A + B = 90o​                       ...(i)

Also, cos (AB) = 1
⇒​ cos (A − B) = cos 0o                    [∵​ cos 0o = 1]
A B = 0o                        ...(ii)

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

Page No 303:

Question 24:

If sin (AB) = 12 and cos (A + B) = 12, 0° < (A + B) < 90° and A > B, then find A and B.

Answer:

Here, sin (A − B) = 12 
⇒ sin (A B) = sin 30o                [∵ sin 30o = 12]
A − B = 30o​                                  ...(i)

Also, cos (A + B) = 12
⇒​ cos (A + B) =  cos 60o              [∵​ cos 60o = 12]
A + B = 60o                                 ...(ii)

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

Page No 303:

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 303:

Question 26:

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 303:

Question 27:

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°.



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