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Post a LessonAnswered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-I: Sets and Functions/Trigonometric Functions
Nazia Khanum
Sure, let's prove the trigonometric identity using the properties of triangles.
In triangle ABC, let's denote the sides opposite the angles A, B, and C as a, b, and c respectively. Also, let's denote the angles of the triangle as A, B, and C.
Now, using the sine rule in triangle ABC, we have:
asinA=bsinB=csinCsinAa=sinBb=sinCc
From this, we can express each side of the triangle in terms of the sines of the angles:
a=sinA⋅ka=sinA⋅k b=sinB⋅kb=sinB⋅k c=sinC⋅kc=sinC⋅k
Where kk is a constant.
Now, let's express each term of the given expression in terms of a,b,ca,b,c:
asin(B−C)+bsin(C−A)+csin(A−B)asin(B−C)+bsin(C−A)+csin(A−B)
=(sinA⋅k)sin(B−C)+(sinB⋅k)sin(C−A)+(sinC⋅k)sin(A−B)=(sinA⋅k)sin(B−C)+(sinB⋅k)sin(C−A)+(sinC⋅k)sin(A−B)
=(sinA⋅k)⋅(sinB⋅cosC−cosB⋅sinC)+(sinB⋅k)⋅(sinC⋅cosA−cosC⋅sinA)+(sinC⋅k)⋅(sinA⋅cosB−cosA⋅sinB)=(sinA⋅k)⋅(sinB⋅cosC−cosB⋅sinC)+(sinB⋅k)⋅(sinC⋅cosA−cosC⋅sinA)+(sinC⋅k)⋅(sinA⋅cosB−cosA⋅sinB)
=k(sinA⋅sinB⋅cosC−sinA⋅cosB⋅sinC+sinB⋅sinC⋅cosA−sinB⋅cosC⋅sinA+sinC⋅sinA⋅cosB−sinC⋅cosA⋅sinB)=k(sinA⋅sinB⋅cosC−sinA⋅cosB⋅sinC+sinB⋅sinC⋅cosA−sinB⋅cosC⋅sinA+sinC⋅sinA⋅cosB−sinC⋅cosA⋅sinB)
=k(sinA⋅sinB⋅cosC−sinA⋅cosB⋅sinC+sinB⋅sinC⋅cosA−sinB⋅cosC⋅sinA+sinC⋅sinA⋅cosB−sinC⋅cosA⋅sinB)=k(sinA⋅sinB⋅cosC−sinA⋅cosB⋅sinC+sinB⋅sinC⋅cosA−sinB⋅cosC⋅sinA+sinC⋅sinA⋅cosB−sinC⋅cosA⋅sinB)
=k(sinA⋅sinB⋅cosC+sinB⋅sinC⋅cosA+sinC⋅sinA⋅cosB−sinA⋅cosB⋅sinC−sinB⋅cosC⋅sinA−sinC⋅cosA⋅sinB)=k(sinA⋅sinB⋅cosC+sinB⋅sinC⋅cosA+sinC⋅sinA⋅cosB−sinA⋅cosB⋅sinC−sinB⋅cosC⋅sinA−sinC⋅cosA⋅sinB)
=k(sinA⋅sinB⋅cosC+sinB⋅sinC⋅cosA+sinC⋅sinA⋅cosB−sinA⋅sinB⋅cosC−sinB⋅sinC⋅cosA−sinC⋅sinA⋅cosB)=k(sinA⋅sinB⋅cosC+sinB⋅sinC⋅cosA+sinC⋅sinA⋅cosB−sinA⋅sinB⋅cosC−sinB⋅sinC⋅cosA−sinC⋅sinA⋅cosB)
=0=0
Hence, we have proved that asin(B−C)+bsin(C−A)+csin(A−B)=0asin(B−C)+bsin(C−A)+csin(A−B)=0 in any triangle ABC.
Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-I: Sets and Functions/Trigonometric Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can guide you through solving this geometry problem. UrbanPro is indeed one of the best platforms for finding online coaching and tuition.
To find the radius of the circle when a central angle of 60° intercepts an arc of length 37.4 cm, we'll use a fundamental relationship between the central angle, arc length, and radius of a circle.
The formula to relate the central angle (θθ), arc length (ss), and radius (rr) of a circle is:
s=r⋅θs=r⋅θ
Here, s=37.4s=37.4 cm and θ=60∘θ=60∘. We need to find rr.
First, we need to convert the central angle from degrees to radians because the formula requires angles in radians. Recall that 1∘=π1801∘=180π radians.
So, 60∘=60×π180=π360∘=60×180π=3π radians.
Now, we can rearrange the formula to solve for rr:
r=sθr=θs
Substituting the given values:
r=37.4π3r=3π37.4
To simplify, we'll divide 37.4 by π33π:
r=37.4×3πr=π37.4×3
r=112.2πr=π112.2
Now, let's approximate ππ as 227722:
r≈112.2227r≈722112.2
r≈112.2×722r≈22112.2×7
r≈785.422r≈22785.4
r≈35.7r≈35.7
So, the radius of the circle is approximately 35.735.7 cm.
Remember, UrbanPro is an excellent resource for finding quality tutoring and coaching assistance, whether it's for geometry or any other subject!
Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-I: Sets and Functions/Trigonometric Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your question.
Firstly, to find out how many radians the wheel turns in one second, we need to understand the relationship between revolutions and radians.
One revolution is equivalent to 2π2π radians. So, if the wheel makes 360 revolutions in one minute, it means it covers 360×2π360×2π radians in one minute.
To convert this into radians per second, we divide by 60 (since there are 60 seconds in a minute):
360×2π60=360π60=6π radians per second60360×2π=60360π=6π radians per second
So, the wheel turns 6π6π radians in one second.
If you're looking for further assistance or guidance in mathematics or any other subject, feel free to reach out. I'm here to help!
Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-I: Sets and Functions/Trigonometric Functions
Nazia Khanum
Sure, let's solve this trigonometric expression: √3 cosec 20° – sec 20°.
As an experienced tutor registered on UrbanPro, I'd like to guide you through solving this expression step by step.
First, let's recall the trigonometric identities:
Given that we're dealing with 20°, we need to find the values of sin 20° and cos 20°.
To do this, it's useful to refer to a unit circle or a calculator. We find that sin 20° ≈ 0.342 and cos 20° ≈ 0.940.
Now, we'll substitute these values into our expression:
√3 cosec 20° – sec 20° = √3 * (1/sin 20°) - (1/cos 20°) = √3 * (1/0.342) - (1/0.940) = √3 * 2.920 - 1.064
Now, let's compute these values: √3 * 2.920 ≈ 5.060 1.064 ≈ 1.064
So, the expression simplifies to approximately: 5.060 - 1.064 ≈ 3.996
Therefore, the value of √3 cosec 20° – sec 20° is approximately 3.996.
Remember, if you need further assistance with math or any other subjects, UrbanPro is one of the best online coaching platforms where you can find experienced tutors like myself to help you succeed!
Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-I: Sets and Functions/Trigonometric Functions
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently explain how to solve the trigonometric equation you've presented. First off, UrbanPro is indeed a fantastic platform for online coaching and tuition, offering a wide array of subjects and expert tutors.
Now, let's delve into the problem at hand:
We're tasked with proving the trigonometric identity:
tan(3x) * tan(2x) * tan(x) = tan(3x) - tan(2x) - tan(x)
To demonstrate this identity, we'll employ some fundamental trigonometric identities and algebraic manipulations:
Start with the left side of the equation:
tan(3x) * tan(2x) * tan(x)
Now, let's express tan(3x) in terms of tan(2x) and tan(x) using the tangent addition formula:
tan(3x) = (tan(2x) + tan(x)) / (1 - tan(2x) * tan(x))
Substitute this expression for tan(3x) into the equation:
((tan(2x) + tan(x)) / (1 - tan(2x) * tan(x))) * tan(2x) * tan(x)
Expand and simplify:
(tan(2x) * tan(2x) * tan(x) + tan(x) * tan(2x) * tan(x)) / (1 - tan(2x) * tan(x))
Rewrite tan(2x) * tan(2x) as tan^2(2x):
(tan^2(2x) * tan(x) + tan(x) * tan(2x) * tan(x)) / (1 - tan(2x) * tan(x))
Further simplify:
(tan^2(2x) * tan(x) + tan^2(x) * tan(2x)) / (1 - tan(2x) * tan(x))
Now, recall the identity: tan^2(a) = sec^2(a) - 1
Substitute tan^2(2x) and tan^2(x) accordingly:
((sec^2(2x) - 1) * tan(x) + (sec^2(x) - 1) * tan(2x)) / (1 - tan(2x) * tan(x))
Next, utilize the identity: sec(a) = 1 / cos(a) to replace sec^2(2x) and sec^2(x):
(((1 / cos(2x))^2 - 1) * tan(x) + ((1 / cos(x))^2 - 1) * tan(2x)) / (1 - tan(2x) * tan(x))
Simplify further:
((1 / cos^2(2x) - 1) * tan(x) + (1 / cos^2(x) - 1) * tan(2x)) / (1 - tan(2x) * tan(x))
Yet again, use the identity: cos^2(a) = 1 - sin^2(a) to rewrite the expressions:
(((1 - sin^2(2x)) / cos^2(2x) - 1) * tan(x) + ((1 - sin^2(x)) / cos^2(x) - 1) * tan(2x)) / (1 - tan(2x) * tan(x))
(((cos^2(2x) - sin^2(2x)) / cos^2(2x)) * tan(x) + ((cos^2(x) - sin^2(x)) / cos^2(x)) * tan(2x)) / (1 - tan(2x) * tan(x))
(((cos(2x) / sin(2x)) * tan(x) + (cos(x) / sin(x)) * tan(2x)) / (1 - tan(2x) * tan(x))
((cos(2x) * tan(x) / sin(2x)) + (cos(x) * tan(2x) / sin(x))) / (1 - tan(2x) * tan(x))
((cos(2x) * sin(x) / (sin(2x) * cos(x))) + (cos(x) * (2sin(x)cos(x)) / sin(x))) / (1 - tan(2x) * tan(x))
((cos(2x) * sin(x) + 2cos(x)sin^2(x)) / (sin(2x) * cos(x))) / (1 - tan(2x) * tan(x))
((cos(2x) * sin(x) + sin(2x) * sin(x)) / (sin(2x) * cos(x))) / (1 - tan(2x) * tan(x))
((cos(2x) * sin(x) + sin(2x) * sin(x)) / (sin(2x) * cos(x))) / (1 - tan(2x) * tan(x))
((sin(x) * (cos(2x) + sin(2x))) / (sin(2x) * cos(x))) / (1 - tan(2x) * tan(x))
((cos(2x) + sin(2x)) / cos(x)) / (1 - tan(2x) * tan(x))
((cos(2x) + sin(2x)) / cos(x)) / (1 - (sin(2x) / cos(2x)) * (sin(x) / cos(x)))
((cos(2x) + sin(2x)) / cos(x)) / (cos(2x) * cos(x) - sin(2x) * sin(x)) / (cos(2x) * cos(x))
((cos(2x) + sin(2x)) / cos(x)) / ((cos^2(2x) - sin^2(2x)) * cos(x))
((cos(2x) + sin(2x)) / cos(x)) / ((cos^2(2x) * cos(x)) - (sin^2(2x) * cos(x)))
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