Trigonometric Ratios
Class 10th Mathematics Rajasthan Board Solution
- 2 sin 45° . cos 45° Find the value of the following:
- cos 45° cos 60° - sin 45° sin 60° Find the value of the following:…
- sin^2 30° + 2 cos^2 45° + 3 tan^2 60° Find the value of the following:…
- 3 sin 60° - 4 sin^3 60° Find the value of the following:
- 5cos^260^circle + 4sec^230^circle - tan^245^circle /sin^230^circle +…
- 4cot^2 45° - sec^2 60° + sin^2 60° + cos^2 90° Find the value of the following:…
- 4/cot^230^circle + 1/sin^230^circle - cos^245^circle Find the value of the…
- tan^260^circle + 4sin^245^circle + sin^290^circle /3sec^230^circle +…
- sin30^circle - sin90^circle + 2cos0^circle /tan30^circle tan60^circle Find the…
- 2tan30^circle /1-tan^230^circle Find the value of the following:
- cos x = cos 60° cos 30° + sin 60° sin 30° Find the value of x in the…
- sin 2x = sin 60° cos 30° - cos 60° - sin 30° Find the value of x in the…
- √3 tan 2x = sin 30° + sin 45° cos 45° + 2 sin 90° Find the value of x in the…
- cos30^circle + sin60^circle /1+cos60^circle + sin30^circle = root 3/2 Prove…
- 4cot^245^circle - sec^260^circle - sin^230^circle = - 1/4 Prove that-…
- sin 30° tan^2 60° + 3cos 60° tan 45° = 2sec^2 45° - cosec^2 90° Prove that-…
- cosec^245^circle sec^230^circle sin^390^circle cos60^circle = 4/3 Prove that-…
- sin60^circle + sin30^circle /sin60^circle - sin30^circle = tan60^circle +…
- 2(cos^2 45° + tan^2 60°) - 6(sin^2 45° - tan^2 30°) = 6 Prove that-…
- (sec^2 30° + cosec^2 45°) (2cos 60° + sin 90° + tan 45°) = 10 Prove that-…
- (1-sin45^circle + sin30^circle) (1+cos45^circle + cos60^circle) = 7/4 Prove…
- cos^2 0° - 2 cot^2 30° + 3 cosec^2 90° = 2(sec^2 45° - tan^2 60°) Prove that-…
- sin3x = 3sinx - 4 sin^3 x If x = 30°, then prove that
- tan2x = 2tanx/1-tan^2x If x = 30°, then prove that
- sinx = root 1-cos2x/2 If x = 30°, then prove that
- cos3x = 4 cos^3 x - 3cosx If x = 30°, then prove that
- If A = 60° and B = 30° then prove that cot (a-b) = cotacotb+1/cotb-cota…
- The value of tan^2 60° is:A. 3 B. 1/3 C. 1 D. ∞
- The value of 2 sin^2 60° cos60° will be:A. 4/3 B. 5/2 C. 3/4 D. 1/3…
- If cosectheta = 2/root 3 then the value of θ is:A. pi /4 B. pi /3 C. pi /2 D. pi…
- The value of cos^2 45° will be:A. 1/root 2 B. root 3/2 C. 1/2 D. 1/root 3…
- If θ = 45° then the value of 1-cos2theta /sin2theta is:A. 0 B. 1 C. 2 D. ∞…
- cos60° = 2 cos^2 30° - 1 Prove that:
- sin60^circle = 2tan30^circle /1+tan^230^circle Prove that:
- cos60^circle = 1-tan^230^circle /1+tan^230^circle Prove that:
- (sin45° + cos45°)^2 = 2 Prove that:
- 4 tan30° sin45° sin60° sin90° = √2 Prove that:
- Find the value of sin^2 60° cot^2 60°.
- Find the value of 4cos^3 30° - 3 cos30°.
- If cot θ = 1/root 3 then prove that 1-cos^2theta /2-sin^2theta = 3/5…
- Prove that 3(tan^2 30° + cot^2 30°) - 8(sin^2 45° + cos^2 30°) = 0…
- Prove that 4(sin^4 30° + cos60°) - 3(cos^2 45° - sin^2 90°) = 15/4…
- Prove that cos30^circle + sin60^circle /1+cos60^circle + sin30^circle = root…
- Prove that 2(cos^2 45° + tan^2 60°) - 6(sin^2 45° - tan^2 30°) = 6…
Exercise 6.1
Question 1.Find the value of the following:
2 sin 45° . cos 45°
Answer:
∴ 2 sin 45° .cos 45°
= 
= 1
Question 2.
Find the value of the following:
cos 45° cos 60° – sin 45° sin 60°
Answer:
∴cos 45° cos 60° - sin 45° sin 60°
Question 3.
Find the value of the following:
sin230° + 2 cos245° + 3 tan260°
Answer:
tan60° = √3
∴
∴ sin230° + 2cos245° + 3tan260°
Question 4.
Find the value of the following:
3 sin 60° – 4 sin360°
Answer:
Let θ = 60°
then,
3 sin 60° – 4 sin360°
= 3sinθ - 4sin3θ
We know that
Sin3θ = 3sinθ - 4sin3θ
⇒ 3 sin 60° – 4 sin360°
= sin(3 × 60° )
= sin 180°
= 0
Question 5.
Find the value of the following:
Answer:
tan45° = 1
∴ tan245° = 1
∴
= 
Question 6.
Find the value of the following:
4cot245° – sec260° + sin260° + cos290°
Answer:
cot45° = 1
∴ cot245° = 1
sec60° = 2
∴ sec260° = 4
cos90° = 0
∴ cos290° = 0
∴4cot245° – sec260° + sin260° + cos290°
Question 7.
Find the value of the following:
Answer:
cot30° = √3
∴ cot230° = 3
∴
Question 8.
Find the value of the following:
Answer:
tan 60° = √3
tan260° = 3
sin90° = 1
∴ sin290° = 1
Question 9.
Find the value of the following:
Answer:
sin90° = 1
cos0° = 1
tan60° = √3
∴
= 
Question 10.
Find the value of the following:
Answer:
∴
= 
= 
= √3
Question 11.
Find the value of x in the following:
cos x = cos 60° cos 30° + sin 60° sin 30°
Answer:
We know that,
Cos(A - B) = cosAcosB + sinAsinB
Let A = 60°, B = 30°
Then,
cos 60° cos 30° + sin 60° sin 30°
= cos(A - B)
= cos(60° - 30° )
= cos30°
∴cos x = cos 60° cos 30° + sin 60° sin 30°
⇒ cos x = cos 30°
⇒ x = 30°
Question 12.
Find the value of x in the following:
sin 2x = sin 60° cos 30° – cos 60° – sin 30°
Answer:
We know that,
sin(A - B) = sinAcosB - cosAsinB
LetA = 60°, B = 30°
Then,
sin 60° cos 30° – cos 60°sin 30°
= sin(60° - 30° )
= sin30°
∴sin 2x = sin 60° cos 30° – cos 60°sin 30°
⇒ sin 2x = sin30°
⇒ 2x = 30°
⇒ x = 15°
Question 13.
Find the value of x in the following:
√3 tan 2x = sin 30° + sin 45° cos 45° + 2 sin 90°
Answer:
sin90° = 1
∴sin 30° + sin 45° cos 45° + 2 sin 90°
= 3
∴
tan 2x = sin 30° + sin 45° cos 45° + 2 sin 90°
⇒ √3tan2x = 3
⇒ tan2x = √3
But, tan60° = √3
⇒ tan2x = tan60°
⇒ 2x = 60°
⇒ 
⇒ x = 30°
Question 14.
Prove that-
Answer:
∴LHS =
= 
RHS = 
∴ LHS = RHS
∴
Hence Proved
Question 15.
Prove that-
Answer:
Cot45° = 1
∴ cot245° = 1
Sec60° = 2
∴ sec260° = 4
∴ LHS = 4cot245° - sec260° - sin230°
RHS = 
∴ LHS = RHS
∴
Hence Proved
Question 16.
Prove that-
sin 30° tan260° + 3cos 60° tan 45° = 2sec245° – cosec290°
Answer:
tan45° = 1
sec45° = √2
∴ sec245° = 2
cosec90° = 1
∴ cosec290° = 1
∴ LHS = sin30° tan260° + 3cos60°tan45°
= 
RHS
2sec245° – cosec290°
= 2(2) – 1
= 3
RHS = LHS
Henc, Proved!
Question 17.
Prove that-
Answer:
cosec45° = √2
∴ cosec245° = 2
∴
sin90° = 1
∴ sin390° = 1
∴ LHS = cosec245° .sec230° sin390° cos60°
= RHS
∴ LHS = RHS
∴
Hence Proved
Question 18.
Prove that-
Answer:
tan60° = √3
tan45° = 1
LHS =
(sin60° + sin30° )/(sin60° - sin30° )
RHS = 
= 
∴ LHS = RHS
∴
Hence Proved
Question 19.
Prove that-
2(cos245° + tan260°) – 6(sin245° – tan230°) = 6
Answer:
tan60° = √3
∴ tan260° = 3
∴ LHS =
= 7 - 1
= 6
= RHS
∴ LHS = RHS
∴(cos245° + tan260°) – 6(sin245° – tan230°) = 6
Hence Proved
Question 20.
Prove that-
(sec230° + cosec245°) (2cos 60° + sin 90° + tan 45°) = 10
Answer:
cosec45° = √2
∴ cosec245° = 2
sin90° = 1
tan45° = 1
∴ LHS = (sec230° + cosec245°) (2cos 60° + sin 90° + tan 45°)
= 10 = RHS
∴ LHS = RHS
∴(sec230° + cosec245°) (2cos 60° + sin 90° + tan 45°) = 10
Hence Proved
Question 21.
Prove that-
Answer:
∴LHS = 
= RHS
∴ LHS = RHS
∴
Hence, Proved
Question 22.
Prove that-
cos20° – 2 cot230° + 3 cosec290° = 2(sec245° – tan260°)
Answer:
cos0° = 1
∴ cos20° = 1
cot30° = √3
∴ cot230° = 3
cosec90° = 1
∴ cosec290° = 1
sec45° = √2
∴ sec245° = 2
tan60° = √3
tan260° = 3
∴ LHS = cos20° – 2 cot230° + 3 cosec290°
= 1 - 2 × 3 + 3 × 1
= 1 - 6 + 3
= 4 - 6
= - 2
RHS = 2(sec245° – tan260°)
= 2 × (2 - 3)
= 2 × ( -1)
= - 2
- 2 = - 2
∴ LHS = RHS
∴cos20° – 2 cot230° + 3 cosec290° = 2(sec245° – tan260°)
Hence Proved
Question 23.
If x = 30°, then prove that
sin3x = 3sinx – 4 sin3x
Answer:
For x= 30°,
LHS = sin (3×30°)
= sin (90°)
=1
RHS = 3sin30° - 4× (sin30°)3
Since LHS = RHS
sin3x = 3sinx – 4 sin3x
Hence proved.
Question 24.
If x = 30°, then prove that
Answer:
For, x = 30°,
LHS = tan2x = tan(2 × 30°) = tan60° = √3
RHS =
√3 = √3
∴ LHS = RHS
Hence Proved
Question 25.
If x = 30°, then prove that
Answer:
For x = 30°
LHS =
RHS =
∴ LHS = RHS
∴
Hence, Proved
Question 26.
If x = 30°, then prove that
cos3x = 4 cos3x – 3cosx
Answer:
For x = 30°
LHS = cos(3 × 30° ) = cos90° = 0
RHS = 4 cos3x – 3cosx
= 4 cos330°– 3 cos30°
∴4 cos330°– 3 cos30°
= 
= 0
Since, 0 = 0
∴ LHS = RHS
∴cos3x = 4 cos3x – 3cosx
Hence Proved
Question 27.
If A = 60° and B = 30° then prove that 
Answer:
For A = 60° and B = 30°
LHS = cot(A - B) = cot(60° - 30° ) = cot30° = √3
RHS = 
= √3
√3 = √3
∴ LHS = RHS
∴
Hence Proved
Miscellaneous Exercise 6
Question 1.The value of tan2 60° is:
A. 3
B. 
C. 1
D. ∞
Answer:
tan60° = √3
∴ tan260° = (tan60°)2 = (√3)2 = 3
Question 2.
The value of 2 sin260° cos60° will be:
A. 
B. 
C. 
D. 
Answer:
∴2 sin260° cos60°
Question 3.
If 
then the value of θ is:
A. 
B. 
C. 
D. 
Answer:
Question 4.
The value of cos245° will be:
A. 
B. 
C. 
D. 
Answer:
Question 5.
If θ = 45° then the value of 
is:
A. 0
B. 1
C. 2
D. ∞
Answer:
For θ = 45°
cos2θ = cos(2 × 45° ) = cos90° = 0
sin2θ = sin(2 × 45° ) = sin90° = 1
= 1
Question 6.
Prove that:
cos60° = 2 cos230° – 1
Answer:
LHS =
RHS =
2cos230° - 1
∴ LHS = RHS
∴cos60° = 2 cos230° – 1
Hence Proved
Question 7.
Prove that:
Answer:
∴ LHS
= sin60°
RHS =
∴ LHS = RHS
∴
∴ Hence Proved
Question 8.
Prove that:
Answer:
∴ LHS =
cos60°
RHS =
∴ LHS = RHS
Hence Proved
Question 9.
Prove that:
(sin45° + cos45°)2 = 2
Answer:
LHS:
(sin45° + cos45°)2
= (√2)2
= 2
= RHS
Hence Proved!
Question 10.
Prove that:
4 tan30° sin45° sin60° sin90° = √2
Answer:
sin90° = 1
LHS =
4 tan30° sin45° sin60° sin90°
= √2
= RHS
Hence Proved!
Question 11.
Find the value of sin260° cot260°.
Answer:
∴sin260° cot260°
Question 12.
Find the value of 4cos330° – 3 cos30°.
Answer:
For θ = 30°,
4cos330° – 3 cos30° = 4cos3θ– 3cosθ
We know that,
cos3θ = 4cos3θ– 3cosθ
∴ 4cos330° – 3 cos30°
= cos(3 × 30° )
= cos90°
= 0
Question 13.
If cot θ = 
then prove that 
Answer:
∴ LHS =
∴
= RHS
∴ LHS = RHS
Hence Proved
Question 14.
Prove that 3(tan230° + cot230°) – 8(sin245° + cos230°) = 0
Answer:
cot30° = √3
∴ cot230° = 3
∴ LHS =
3(tan230° + cot230°) – 8(sin245° + cos230°)
= 10 - 10
= 0
= RHS
∴ LHS = RHS
Hence Proved
Question 15.
Prove that 4(sin430° + cos60°) – 3(cos245° – sin290°) = 
Answer:
sin90° = 1
sin290° = 1
∴ LHS = 4(sin430° + cos60°) – 3(cos245° – sin290°)
RHS = LHS
Hence Proved
Question 16.
Prove that 
Answer:
∴ LHS =
Question 17.
Prove that 2(cos245° + tan260°) – 6(sin245° – tan230°) = 6
Answer:
∴ LHS = 2(cos245° + tan260°) – 6(sin245° – tan230°)
= 7 – 1
= 6
= RHS
= Hence Proved.