Question

tan A. = 1/root 3 tan B = root 3
tan (A+B) = ?​

Answer 1

Question :-

$\sf{ \tan( {A} ) = \frac{1}{ \sqrt{3} } \: \: \: and \: \tan(B) = \sqrt{3} } \\ \sf{then \:find \: \tan(A+B) }$

Answer :-

$\implies \boxed{\bf{\tan(A+B) = \infty}} \:$

Formula used :-

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

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Solution :-

Given that ,

$\implies \sf{ \tan(A) = \frac{1}{ \sqrt{3} } \: and \: \tan(B) = \sqrt{3} }$

Therefore,

$\rightarrow \sf{\tan(A+B) = \frac{ \frac{1}{ \sqrt{3} } + \sqrt{ 3} }{1 - \frac{1}{ \sqrt{3} \: \: } \: \sqrt{3} } } \\ \\ \implies \sf{\frac{ \frac{1 + 3}{ \sqrt{3} } }{1 - 1} } \\ \\ \implies \: \frac{ \frac{4}{ \sqrt{3} } }{0} \\ \\ \implies \boxed{\bf{\tan(A+B) = \infty}}$

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Answer 2

$\color {red}\huge\bold\star\underline\mathcal{Question:-}$

$\textbf{we have to find value of tan(A+B)}$

$\huge\underline\blue{\sf Given:}$ TanA = 1/√3 , tanB = √3

$\</strong><strong>l</strong><strong>a</strong><strong>r</strong><strong>g</strong><strong>e</strong><strong>\boxed{\fcolorbox{fuchsia}{grey}{Solution</strong><strong> </strong><strong>(</strong><strong>1</strong><strong>)</strong><strong>:--}}$

we know that Tan30° = 1/√3 and Tan60° = √3

so,

TanA = 1/√3 = Tan30°

$\large\red{\boxed{\sf </strong><strong>A</strong><strong>=</strong><strong>3</strong><strong>0</strong><strong>°</strong><strong>}}$

TanB = √3 = Tan30°

$\large\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{\boxed{\sf A=</strong><strong>6</strong><strong>0°}}$

so,

Tan(A+B) = Tan(30°+60°) = Tan90°

$\large\</strong><strong>p</strong><strong>i</strong><strong>n</strong><strong>k</strong><strong>{\boxed{\sf </strong><strong>Tan</strong><strong>(</strong><strong>A</strong><strong>+</strong><strong>B</strong><strong>)</strong><strong>\</strong><strong>:</strong><strong>=</strong><strong>\infty</strong><strong>}}$

$\rule{200}{4}$

$\large\boxed{\fcolorbox{fuchsia}{grey}{Solution (</strong><strong>2</strong><strong>):--}}$

we know the formula ,

$\tan(a + b) = \frac{tana + tanb}{1 - tana \times tanb}$

$\textbf{</strong><strong>Putting</strong><strong> </strong><strong>Given</strong><strong> </strong><strong>values</strong><strong> </strong><strong>here</strong><strong> </strong><strong>we</strong><strong> </strong><strong>get</strong><strong>}$

$\tan(a + b) = \frac{ \frac{1}{ \sqrt{3} } + \sqrt{3} }{1 - \frac{1}{ \sqrt{3} } \times \sqrt{3} } \\ \\ \\ \\ \tan(a + b) = \frac{ \frac{(1 + 3)}{ \sqrt{3} } }{1 - 1} \\ \\ \\ \tan(a + b) = \frac{ \frac{4}{ \sqrt{3} } }{0} \\ \\ \\ \tan(a + b) = \infty$

so,

$\large\</strong><strong>b</strong><strong>l</strong><strong>u</strong><strong>e</strong><strong>{\boxed{\sf Tan(A+B)\:=\infty}}$

$\huge\underline\mathfrak\green{Hope\:it\:Helps\:You}$