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Standard Values of Trigonometric Ratios

If A and B ar...

Question

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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Solution

given

$tan space space left parenthesis space A space plus space B space right parenthesis space equals space fraction numerator tan space A thin space plus tan space B thin space over denominator 1 minus tan space A thin space asterisk times space tan space B end fraction tan space left parenthesis A plus B right parenthesis equals fraction numerator begin display style 1 half end style plus begin display style 1 third end style over denominator 1 minus begin display style fraction numerator 1 over denominator 2 asterisk times 3 end fraction end style end fraction tan space space left parenthesis A plus B right parenthesis equals fraction numerator begin display style fraction numerator 3 plus 2 over denominator 6 end fraction end style over denominator begin display style fraction numerator 6 minus 1 over denominator 6 end fraction end style end fraction tan space left parenthesis A plus B right parenthesis equals fraction numerator begin display style 5 over 6 end style over denominator begin display style 5 over 6 end style end fraction tan space left parenthesis A plus B right parenthesis equals 1 s o space A plus B equals 45$

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Similar questions

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

\frac{1}{3}

\frac{1}{2}

and tan (A + B) =

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

, show that A + B = 45°.

A

B

are obtuse angles such that,

tan A + tan B + tan A tan B = 1

A - B = 41^{\circ}

A = 60^{o}

B = 30^{o}

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

A = 60^{o}

B = 30^{o}

, then prove that

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

∠ A

∠ B

are acute angles such that tanA = tanB, the prove that

∠ A = ∠ B

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