Question

$If \: cos \: A \: = \: \dfrac{1}{2} \: and \: sin \: B \: = \: \dfrac{1}{ \sqrt{2} } \: ,find \: the \: value \: of \: : \\ \frac{tan \:A \: - \: tan \:B}{1 + tan \:A \: tan \: B \: .} \\ Here \: angles \: A \: and \: B \: are \: from \: same \: triange$
​

Answer 1

Given that,

$\sf \: tan( \dfrac{ \theta}{2} ) = \sqrt{ \dfrac{a - b}{a + b} } tan( \dfrac{ \phi}{2} )$

To Prove :

$\sf \: cos \: \theta = \dfrac{acos \: \phi + b}{a + bcos \: \phi}$

We Know that,

$\sf \: cos2x = \dfrac{1 - tan {}^{2}x }{1 + tan {}^{2}x }$

Similarly,

$\longrightarrow \sf cos \ \theta = \dfrac{1 - tan^2 \frac{\theta}{2}}{1 + tan^2 \frac{\theta}{2}}$

Then,

$\sf \: cos \ \theta = \dfrac{1 - ( \sqrt{ \dfrac{a - b}{a + b} } \: tan \frac{ \phi}{2} ) {}^{2} }{1 + ( \sqrt{ \dfrac{a - b}{a + b} } \: tan \frac{ \phi}{2}) {}^{2} } \\ \\ \\ \longrightarrow \: \sf \: cos \: \theta = \frac{1 - \bigg( \dfrac{a - b}{a + b} \times \dfrac{ {sin}^{2} \frac{ \phi}{2} }{ {cos}^{2} \frac{ \phi}{2} } \bigg)}{1 + \bigg( \dfrac{a - b}{a + b} \times \dfrac{ {sin}^{2} \frac{ \phi}{2} }{ {cos}^{2} \frac{ \phi}{2} } \bigg) } \\ \\ \\ \longrightarrow \: \sf \: cos \: \theta = \dfrac{(a + b) {cos}^{2} \frac{ \phi}{2} - (a - b) {sin}^{2} \frac{ \phi}{2} }{(a + b) {cos}^{2} \frac{ \phi}{2} + (a - b) {sin}^{2} \frac{ \phi}{2} } \\ \\ \\ \longrightarrow \: \sf \: cos \: \theta = \dfrac{a(cos {}^{2} \frac{ \phi}{2} - {sin}^{2} \frac{ \phi}{2}) + b(cos {}^{2} \frac{ \phi}{2} + {sin}^{2} \frac{ \phi}{2} ) }{a(cos {}^{2} \frac{ \phi}{2} + {sin}^{2} \frac{ \phi}{2} ) + b(cos {}^{2} \frac{ \phi}{2} - {sin}^{2} \frac{ \phi}{2} ) } \\ \\ \\ \longrightarrow \: \sf \: cos \: \theta = \frac{acos \: \phi + b}{a + bcos \: \phi}$

$\bf\blue{ Hence~proved}$

$\bf\green{ Identities~used}$

cos²∅ - sin²∅ = cos2∅

sin²∅ + cos²∅ = 1

Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{cosA = \dfrac{1}{2} } \\ &\sf{sinB = \dfrac{1}{ \sqrt{2} } } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{\dfrac{tanA - tanB}{1 + tanAtanB} }  \end{cases}\end{gathered}\end{gathered}$

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

$\tt \:  \longrightarrow \: cosA = \dfrac{1}{2}$

$\tt \:  \longrightarrow \: cosA = cos60 \degree$

$\tt\implies \:A \: = \: 60 \degree$

$\tt \:  \longrightarrow \: sinB = \dfrac{1}{ \sqrt{2} }$

$\tt \:  \longrightarrow \: sinB = sin45 \degree$

$\tt\implies \:B \: = \: 45 \degree$

$\begin{gathered}\begin{gathered}\bf So = \begin{cases} &\sf{A = 60 \degree} \\ &\sf{B = 45 \degree} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\bf\red{Now, \: consider}\end{gathered}$

$\tt \:  \longrightarrow \: \dfrac{tanA - tanB}{1 + tanAtanB}$

$\tt \:  \longrightarrow \: = \dfrac{tan60 \degree - tan45 \degree}{1 + tan60 \degree \: tan45 \degree}$

$\tt \:  \longrightarrow \: = \dfrac{ \sqrt{3} - 1 }{1 + \sqrt{3} \times 1 }$

$\tt \:  \longrightarrow \: = \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } \times \dfrac{ \sqrt{3} - 1}{ \sqrt{3} - 1 }$

$\tt \:  \longrightarrow \: = \dfrac{ {( \sqrt{3} - 1)}^{2} }{ {( \sqrt{3} )}^{2} - {(1)}^{2} }$

$\tt \:  \longrightarrow \: = \dfrac{3 + 1 - 2 \sqrt{3} }{3 - 1}$

$\tt \:  \longrightarrow \: = \dfrac{4 - 2 \sqrt{3} }{2}$

$\tt \:  \longrightarrow \: = \dfrac{2(2 - \sqrt{3} )}{2}$

$\tt \:  \longrightarrow \: = 2 - \sqrt{ 3}$

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$\large \red{\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Explore \: more }$

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