Tan A-tan B A- B-tan A tan B

Simplifying the LHS of tan A + tan B A + B = tan Atan B . tan A + tan B A + B = sin Acos A + sin Bcos Bcos Asin A + cos Bsin B = sin Acos B + co ...

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Standard XII

Mathematics

Derivative of Standard Functions

tan A+tan B A...

Question

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

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Solution

Simplifying the LHS of $\frac{tan A + tan B}{cot A + cot B} = tan A tan B$ .

$\frac{tan A + tan B}{cot A + cot B} = \frac{\frac{sin A}{cos A} + \frac{sin B}{cos B}}{\frac{cos A}{sin A} + \frac{cos B}{sin B}}$

$= \frac{\frac{sin A cos B + cos A sin B}{cos A cos B}}{\frac{cos A sin B + sin A cos B}{sin A sin B}}$

$= \frac{sin (A + B) sin A sin B}{sin (A + B) cos A cos B}$

$= tan A tan B$

$= R H S$

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Similar questions

Q. Prove that

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

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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

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tanA+tanBcotA+cotB=tanAtanB

Simplifying the LHS of tanA+tanBcotA+cotB=tanAtanB.tanA+tanBcotA+cotB=sinAcosA+sinBcosBcosAsinA+cosBsinB=sinAcosB+cosAsinBcosAcosBcosAsinB+sinAcosBsinAsinB=sin(A+B)sinAsinBsin(A+B)cosAcosB=tanAtanB=RHS

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