Math Problem Statement

find tanB in terms of tanA if tanA= (cos(A+B))/(cos(A-B))

Solution

To find tanB\tan B in terms of tanA\tan A, given that tanA=cos(A+B)cos(AB)\tan A = \frac{\cos(A+B)}{\cos(A-B)}, we can use trigonometric identities and relationships.

Given:

tanA=cos(A+B)cos(AB)\tan A = \frac{\cos(A + B)}{\cos(A - B)}

We will use the sum and difference formulas for cosine:

cos(A+B)=cosAcosBsinAsinB\cos(A + B) = \cos A \cos B - \sin A \sin B

cos(AB)=cosAcosB+sinAsinB\cos(A - B) = \cos A \cos B + \sin A \sin B

Substitute these into the equation for tanA\tan A:

tanA=cosAcosBsinAsinBcosAcosB+sinAsinB\tan A = \frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B + \sin A \sin B}

Divide both the numerator and the denominator by cosAcosB\cos A \cos B:

tanA=cosAcosBsinAsinBcosAcosBcosAcosB+sinAsinBcosAcosB\tan A = \frac{\frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B}}{\frac{\cos A \cos B + \sin A \sin B}{\cos A \cos B}}

Simplify by separating the fractions:

tanA=1sinAsinBcosAcosB1+sinAsinBcosAcosB\tan A = \frac{1 - \frac{\sin A \sin B}{\cos A \cos B}}{1 + \frac{\sin A \sin B}{\cos A \cos B}}

We can use the identity sinX=tanXcosX\sin X = \tan X \cos X to rewrite the fractions:

tanA=1tanAtanBcosAcosB1+tanAtanBcosAcosB\tan A = \frac{1 - \frac{\tan A \tan B}{\cos A \cos B}}{1 + \frac{\tan A \tan B}{\cos A \cos B}}

Simplify further by observing that tanAtanBcosAcosB=tanAtanB\frac{\tan A \tan B}{\cos A \cos B} = \tan A \tan B:

tanA=1tanAtanB1+tanAtanB\tan A = \frac{1 - \tan A \tan B}{1 + \tan A \tan B}

Rearrange this equation to isolate tanB\tan B:

tanA(1+tanAtanB)=1tanAtanB\tan A (1 + \tan A \tan B) = 1 - \tan A \tan B

Expand and rearrange to get terms involving tanB\tan B on one side:

tanA+tanA2tanB=1tanAtanB\tan A + \tan A^2 \tan B = 1 - \tan A \tan B

tanA2tanB+tanAtanB=1tanA\tan A^2 \tan B + \tan A \tan B = 1 - \tan A

Factor out tanB\tan B from the left-hand side:

tanB(tanA2+tanA)=1tanA\tan B (\tan A^2 + \tan A) = 1 - \tan A

Solve for tanB\tan B:

tanB=1tanAtanA2+tanA\tan B = \frac{1 - \tan A}{\tan A^2 + \tan A}

Thus, tanB\tan B in terms of tanA\tan A is:

tanB=1tanAtanA2+tanA\tan B = \frac{1 - \tan A}{\tan A^2 + \tan A}

Would you like more details or have any questions about this solution?

Here are 5 related questions to expand your understanding:

  1. What is the value of tanB\tan B if tanA=1\tan A = 1?
  2. How would you find tan(A+B)\tan(A + B) in terms of tanA\tan A and tanB\tan B?
  3. Can you derive an expression for sinB\sin B in terms of sinA\sin A and cosA\cos A?
  4. How do the sum and difference formulas for sine and cosine relate to the tangent formula used here?
  5. What happens to the value of tanB\tan B as tanA\tan A approaches infinity?

Tip: To simplify trigonometric expressions, always consider using fundamental identities like the Pythagorean identities and sum/difference formulas.

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Math Problem Analysis

Mathematical Concepts

Trigonometry

Formulas

Sum and difference formulas for cosine
Trigonometric identities

Theorems

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Suitable Grade Level

Advanced High School