Solve tanA - B - C - -tan A -tan A1 - -tan A tan B

tan(2 π5 π15) tan (A + B + C) = ∑tan A . πtan A1 ∑tan A. tan B tan (A + B + C) = (tan A + tan B + tan C) tan A. tan B . tan C1 (tan A. tan ...

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Byju's Answer

Standard XII

Mathematics

Circular Measurement of Angle

Solve tanA ...

Question

Solve
$tan (A + B + C) = \frac{\sum tan A \prod tan A}{1 - \sum tan A tan B}$

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Solution

$tan (\frac{2 π}{5} - \frac{π}{15})$
$tan (A + B + C) = \frac{\sum tan A . π tan A}{1 - \sum tan A . tan B}$
$tan (A + B + C) = \frac{(tan A + tan B + tan C) tan A . tan B . tan C}{1 - (tan A . tan B + tan B + tan C + tan C . tan A)}$
$\frac{(tan A + tan B + tan C) - tan A . tan B . tan C}{1 - (tan A . tan B + tan C . tan A + tan B . tan C)}$ =
Both are equal Hence proved

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Similar questions

Q. If

A = 60^{o}

and

B = 30^{o}

, verify that

tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}

Q. If

A = 60^{o}

and

B = 30^{o}

, then prove that

tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

If A and B are acute angles such that tan $A = \frac{1}{2}, t a n B = \frac{1}{3}$ and tan $(A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$ ,find A + B

\frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \tan B

Q. Prove the following trigonometric identities.

(i)

\frac{\cot A + \tan B}{\cot B + \tan A} = \cot A \tan B

(ii)

\frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \tan B

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Solve tan(A+B+C)=∑tanA∏tanA1−∑tanAtanB

tan(2π5−π15)tan(A+B+C)=∑tanA.πtanA1−∑tanA.tanBtan(A+B+C)=(tanA+tanB+tanC)tanA.tanB.tanC1−(tanA.tanB+tanB+tanC+tanC.tanA)(tanA+tanB+tanC)−tanA.tanB.tanC1−(tanA.tanB+tanC.tanA+tanB.tanC) = Both are equal Hence proved

Solve
$tan (A + B + C) = \frac{\sum tan A \prod tan A}{1 - \sum tan A tan B}$