Question

hi,
tan alpha=m/m+1 and tan beta=1/2m+1 find alpha - beta
plz ans

Answer 1

Answer:

please refer to the attached

Answer 2

Given :-

• $\sf\ \: \alpha \: = \: \frac{m}{m \: + \: 1} \: + \: \: \sf\ \beta \: = \: \frac{1}{2m \: + \: 1}$

To Find :-

• $\sf\ The \: value \: of \: \: \alpha \: + \: \beta \: = \:$

Solution :-

L.H.S = tan ($\alpha \: + \: \beta$)

${\large{⇒}}\sf\ \: \: \dfrac{tan \: \alpha \: + \: tan \: \beta}{1 \: - \: tan \: \alpha \: tan \: \beta}$

${\large{⇒}}\sf\ \: \: \dfrac{ \frac{m}{m \: + \: 1} \: + \: \frac{1}{2m \: + \: 1}}{1 \: - \: ( \frac{m}{m \: + \: 1}) \: \frac{1}{2m \: + \: 1}}$

${\large{⇒}}\sf\ \: \: \dfrac{\frac{m \: (2n \: + \: 1) \: + \: m \: + \: 1}{\cancel{(m \: + \: 1) \: (2m \: + \: 1)}}}{\frac{(m \: + \: 1) \: (2m \: + \: 1) \: - \: m}{{\cancel{(m \: + \: 1) \: (2m \: + \: 1)}}}}$

${\large{⇒}}\sf\ \: \: \dfrac{{2m}^{2} \: + \: m \: + \: m \: + \: 1}{{2m}^{2} \: {\cancel{ + \: m}} \: + \: {\cancel{2m}} \: + \: 1 \: {\cancel{ - \: m}}}$

${\large{⇒}}\sf\ \: \: \dfrac{2m {}^{2} \: {\cancel{ + \: 2m \: + \: 1}}}{2m {}^{2} \: {\cancel{ + \: m \: + \: 1}}} \: = \: 1$

Now,

• tan ( $\alpha \: + \: \beta$ ) = 1
• tan ( $\alpha \: + \: \beta$ ) = tan 45

Let,

• $\alpha \: + \: \beta$ = 45
• $\alpha \: + \: \beta$ = ${\Large{\sf{\red{\frac{ \pi}{4}}}}}$

$\sf\ The \: Answer \: will \: be \: \large\sf\frac{ \pi}{4}$

$\tiny$

Hence, Proved!

Additional Information :-

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