Question

I will mark you as the brainliest if you answer it correctly ​

Answer 1

Question :–

• If $\rm \: \: \tan(A - B) = \dfrac{1 }{ \sqrt{ 3} } \: \:$ and $\rm \: \: \tan(A + B) = \sqrt{ 3} \: \:$ , then find A and B .

ANSWER :–

GIVEN :–

$\\ \: \: {\huge {.}} \rm \: \: \tan(A - B) = \dfrac{1 }{ \sqrt{ 3} } \: \:$

$\\ \: \: {\huge {.}} \rm \: \: \tan(A + B) = \sqrt{ 3} \: \:$

TO FIND :–

• Value of 'A' and 'B' = ?

SOLUTION :–

• We know that –

$\\ \: \: {\huge {.}} \rm \: \: \tan( {30}^{ \circ} ) = \dfrac{1 }{ \sqrt{ 3} } \: \:$

$\\ \: \: {\huge {.}} \rm \: \: \tan( {60}^{ \circ} ) = \sqrt{ 3} \: \:$

• Now –

$\\ \implies \rm \tan(A - B) = \dfrac{1 }{ \sqrt{ 3} } \: \:$

$\\ \implies \rm \tan(A - B) = \tan( {30}^{ \circ} ) \: \:$

$\\ \implies \rm A - B = {30}^{ \circ} \: \: \: \: \: - - - - eq.(1)$

• And –

$\\ \implies \rm \tan(A + B) = \sqrt{ 3} \: \:$

$\\ \implies \rm \tan(A + B) = \tan( {60}^{ \circ} ) \: \:$

$\\ \implies \rm A + B = {60}^{ \circ} \: \: \: \: \: - - - - eq.(2)$

• Add eq.(1) and eq.(2) –

$\\ \implies \rm A - B + A + B = {30}^{ \circ} + {60}^{ \circ}$

$\\ \implies \rm 2A = {90}^{ \circ}$

$\\ \longrightarrow \: \: \large{ \boxed{ \rm A = {45}^{ \circ} }}$

• Using eq.(2) –

$\\ \implies \rm {45}^{ \circ} + B = {60}^{ \circ}$

$\\ \implies \rm B = {60}^{ \circ} - {45}^{ \circ}$

$\\ \longrightarrow \: \: \large{ \boxed{ \rm B = {15}^{ \circ} }}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\dagger{\sf{ \: \: \: tan(A + B) = \sqrt{3} }}$

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

${\bf{\blue{\underline{To\:Find:}}}}$

$\dagger{\sf{ \: \: \: A \: and \: B}}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ tan(A + B) = \sqrt{3} }}$

$: \implies{\sf{ tan(A + B) = tan60 \degree}}$

$: \implies{\sf{ \cancel{ tan}(A + B) = \cancel{tan}60 \degree }}$

$: \implies{\sf{ A + B= 60 \degree.......(1) }}$

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

$: \implies{\sf{ tan(A -B) = tan30 \degree}}$

$: \implies{\sf{ \cancel{ tan}(A - B) = \cancel{tan}30 \degree }}$

$: \implies{\sf{ A -B= 30 \degree.......(2) }}$

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From (1) and (2),

A + B = 60°

A - B. = 30°

_________

2A. = 90°

_________

$\longrightarrow{\sf{ A = \frac{90 \degree}{2} }}$

$\longrightarrow{\sf{ A = 45 \degree }}$

Put the Value of A in eq (1),

$:\implies{\sf{ A + B= 60 \degree }}$

$:\implies{\sf{ 45 \degree + B= 60 \degree }}$

$:\implies{\sf{ B= 60 \degree-45\degree }}$

$:\implies{\sf{ B= 15\degree }}$

$\dagger \: \: { \boxed{ \sf{ \purple{Hence \: A = 45 \degree \: and \: B= 15 \degree }}}}$