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Popular Trigonometry >

tan(arccos(24/25)-arcsin(12/13))

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Solution

tan(arccos(2524​)−arcsin(1312​))

Solution

−204253​
+1
Decimal
−1.24019…
Solution steps
tan(arccos(2524​)−arcsin(1312​))
Rewrite using trig identities:1+tan(arccos(2524​))tan(arcsin(1312​))tan(arccos(2524​))−tan(arcsin(1312​))​
tan(arccos(2524​)−arcsin(1312​))
Use the Angle Difference identity: tan(s−t)=1+tan(s)tan(t)tan(s)−tan(t)​=1+tan(arccos(2524​))tan(arcsin(1312​))tan(arccos(2524​))−tan(arcsin(1312​))​
=1+tan(arccos(2524​))tan(arcsin(1312​))tan(arccos(2524​))−tan(arcsin(1312​))​
Rewrite using trig identities:tan(arccos(2524​))=247​
tan(arccos(2524​))
Rewrite using trig identities:tan(arccos(2524​))=(2524​)1−(2524​)2​​
Use the following identity: tan(arccos(x))=x1−x2​​
=(2524​)1−(2524​)2​​
=2524​1−(2524​)2​​
Simplify=247​
Rewrite using trig identities:tan(arcsin(1312​))=512​
tan(arcsin(1312​))
Rewrite using trig identities:tan(arcsin(1312​))=1−(1312​)2(1312​)1−(1312​)2​​
Use the following identity: tan(arcsin(x))=1−x2x1−x2​​
=1−(1312​)2(1312​)1−(1312​)2​​
=1−(1312​)21312​1−(1312​)2​​
Simplify=512​
=1+247​⋅512​247​−512​​
Simplify 1+247​⋅512​247​−512​​:−204253​
1+247​⋅512​247​−512​​
247​⋅512​=107​
247​⋅512​
Cross-cancel common factor: 12
12,24
Greatest Common Divisor (GCD)
Prime factorization of 12:2⋅2⋅3
12
12divides by 212=6⋅2=2⋅6
6divides by 26=3⋅2=2⋅2⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅2⋅3
Prime factorization of 24:2⋅2⋅2⋅3
24
24divides by 224=12⋅2=2⋅12
12divides by 212=6⋅2=2⋅2⋅6
6divides by 26=3⋅2=2⋅2⋅2⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅2⋅2⋅3
The prime factors common to 12,24 are =2⋅2⋅3
Multiply the numbers: 2⋅2⋅3=12=12
=27​⋅51​
Multiply fractions: ba​⋅dc​=b⋅da⋅c​=2⋅57⋅1​
Multiply the numbers: 7⋅1=7=2⋅57​
Multiply the numbers: 2⋅5=10=107​
=1+107​247​−512​​
Join 247​−512​:−120253​
247​−512​
Least Common Multiplier of 24,5:120
24,5
Least Common Multiplier (LCM)
Prime factorization of 24:2⋅2⋅2⋅3
24
24divides by 224=12⋅2=2⋅12
12divides by 212=6⋅2=2⋅2⋅6
6divides by 26=3⋅2=2⋅2⋅2⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅2⋅2⋅3
Prime factorization of 5:5
5
5 is a prime number, therefore no factorization is possible=5
Multiply each factor the greatest number of times it occurs in either 24 or 5=2⋅2⋅2⋅3⋅5
Multiply the numbers: 2⋅2⋅2⋅3⋅5=120=120
Adjust Fractions based on the LCM
Multiply each numerator by the same amount needed to multiply its
corresponding denominator to turn it into the LCM 120
For 247​:multiply the denominator and numerator by 5247​=24⋅57⋅5​=12035​
For 512​:multiply the denominator and numerator by 24512​=5⋅2412⋅24​=120288​
=12035​−120288​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=12035−288​
Subtract the numbers: 35−288=−253=120−253​
Apply the fraction rule: b−a​=−ba​=−120253​
=1+107​−120253​​
Apply the fraction rule: b−a​=−ba​=−1+107​120253​​
Apply the fraction rule: acb​​=c⋅ab​1+107​120253​​=120(1+107​)253​=−120(1+107​)253​
Join 1+107​:1017​
1+107​
Convert element to fraction: 1=101⋅10​=101⋅10​+107​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=101⋅10+7​
1⋅10+7=17
1⋅10+7
Multiply the numbers: 1⋅10=10=10+7
Add the numbers: 10+7=17=17
=1017​
=−120⋅1017​253​
Multiply 120⋅1017​:204
120⋅1017​
Multiply fractions: a⋅cb​=ca⋅b​=1017⋅120​
Multiply the numbers: 17⋅120=2040=102040​
Divide the numbers: 102040​=204=204
=−204253​
=−204253​

Popular Examples

cot(29)3sin((4pi)/3)-arcsin(0)arcsin(sin((10pi)/7))arcsin(tan(-pi/4))

Frequently Asked Questions (FAQ)

  • What is the value of tan(arccos(24/25)-arcsin(12/13)) ?

    The value of tan(arccos(24/25)-arcsin(12/13)) is -253/204
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