Prove that tan A - tan B A - B - tan A tan B

To Prove: tan A+tan Btan A + B=tan A. tan B L.H.S =tan A+tan B1tan A+1tan B=tan A+tan Btan B+tan Atan A.tan B [∵ tan A=1cos A] =tan A.tan ...

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Standard X

Mathematics

Trigonometric Identities

Prove that ...

Question

Prove that $\frac{tan A + tan B}{cot A + cot B} = tan A tan B$

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Solution

To Prove: $\frac{tan A + tan B}{tan A + cot B} = tan A . tan B$
$L . H . S$
$= \frac{tan A + tan B}{\frac{1}{tan A} + \frac{1}{tan B}} = \frac{tan A + tan B}{\frac{tan B + tan A}{tan A . tan B}} [∵ tan A = \frac{1}{cos A}]$
$= tan A . tan B = R . H . S$

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Similar questions

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}

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

Q. If

A = 60^{o}

and

B = 30^{o}

, verify 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

Q. If A and B are acute angles such that tan A =

\frac{1}{3}

, tan B =

\frac{1}{2}

and tan (A + B) =

\frac{\tan A + \tan B}{1 - \tan A \tan B}

, show that A + B = 45°.

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Prove that tanA+tanBcotA+cotB=tanAtanB

To Prove: tanA+tanBtanA+cotB=tanA.tanBL.H.S=tanA+tanB1tanA+1tanB=tanA+tanBtanB+tanAtanA.tanB[∵ tanA=1cosA]=tanA.tanB=R.H.S

Prove that $\frac{tan A + tan B}{cot A + cot B} = tan A tan B$

To Prove: $\frac{tan A + tan B}{tan A + cot B} = tan A . tan B$
$L . H . S$
$= \frac{tan A + tan B}{\frac{1}{tan A} + \frac{1}{tan B}} = \frac{tan A + tan B}{\frac{tan B + tan A}{tan A . tan B}} [∵ tan A = \frac{1}{cos A}]$
$= tan A . tan B = R . H . S$