Question

If tan (A+B) = √3 and tan (A-B ) = 1/√3 . Find A and B.

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Answer 1

EXPLANATION.

⇒ tan(A + B) = √3. - - - - - (1).

⇒ tan(A - B) = 1/√3. - - - - - (2).

As we know that,

We can write equation as, we get.

⇒ tan(A + B) = tan(60°). - - - - - (1).

⇒ tan(A - B) = tan(30°). - - - - - (2).

⇒ A + B = 60°. - - - - - (1).

⇒ A - B = 30°. - - - - - (2).

We get,

⇒ 2A = 90°.

⇒ A = 45°.

Put the value of A = 45° in equation (1), we get.

⇒ 45° + B = 60°.

⇒ B = 60° - 45°.

⇒ B = 15°.

Value of A = 45° & B = 15°.

Answer 2

${\large{\pmb{\sf{\underline{Required \; Solution...}}}}}$

Given that:

• Tan (A+B) = √3
• Tan (A-B) = 1/√3

To find:

• The value of A
• The value of B

Solution:

• The value of A = 45°
• The value of B = 15°

Knowledge required:

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \tt sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \tt cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \tt tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \tt \infty \\ \\ \tt cosec A & \tt \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \tt sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \tt \infty \\ \\ \tt cotA & \tt \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}$

Full Solution:

Given that:

• Tan (A+B) = √3

• Tan (A-B) = 1/√3

According to trigonometry table, we can also write these as:

• Tan (A+B) = √3 as Tan (A+B) = Tan(60°)

• Tan (A-B) = 1/√3 as Tan (A-B) = Tan(30°)

Hence, according to the question,

• A+B = Tan(60°)

• A-B = Tan(30°)

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

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

~ Now according to the equation,

${\sf{:\implies A+A \: = 2A}}$

${\sf{:\implies 60 + 30 = 90 \degree}}$

${\sf{:\implies \cancel{-B \: and \: +B}}}$

~ Now let's find the value of A

${\sf{:\implies 2A \: = 90}}$

${\sf{:\implies A \: = 90/2}}$

${\sf{:\implies A \: = 45 \degree}}$

~ Now finding the value of B

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

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

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

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