Math, asked by talpadadilip417, 20 hours ago

if 2\left|\begin{array}{ll}\tt\sin (A+B) &\tt \cos (A+B) \\ \tt\cos (A-B) & \tt\sin (A-B)\end{array}\right|+\sqrt{3}=0, then find A.

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Answered by mathdude500
18

\large\underline{\sf{Solution-}}

Given determinant is

\rm \: 2\left|\begin{array}{ll}\tt\sin (A+B) &\tt \cos (A+B) \\ \tt\cos (A-B) & \tt\sin (A-B)\end{array}\right|+\sqrt{3}=0 \\

can be rewritten as

\rm \: 2\left|\begin{array}{ll}\tt\sin (A+B) &\tt \cos (A+B) \\ \tt\cos (A-B) & \tt\sin (A-B)\end{array}\right|= -  \sqrt{3}  \\

\rm \: \left|\begin{array}{ll}\tt\sin (A+B) &\tt \cos (A+B) \\ \tt\cos (A-B) & \tt\sin (A-B)\end{array}\right|= -\dfrac{ \sqrt{3} }{2} \\

\rm \: sin(A + B)sin(A - B) - cos(A + B)cos(A - B)= -\dfrac{ \sqrt{3} }{2} \\

\rm \:({sin}^{2}A- {sin}^{2}B)-( {cos}^{2}A-{sin}^{2}B)  = -\dfrac{ \sqrt{3} }{2} \\

\rm \:{sin}^{2}A- {sin}^{2}B-{cos}^{2}A + {sin}^{2}B  = -\dfrac{ \sqrt{3} }{2} \\

\rm \:{sin}^{2}A-{cos}^{2}A  = -\dfrac{ \sqrt{3} }{2} \\

\rm \: - ( - {sin}^{2}A + {cos}^{2}A)  = -\dfrac{ \sqrt{3} }{2} \\

\rm \: {cos}^{2}A - {sin}^{2}A = \dfrac{ \sqrt{3} }{2} \\

\rm \: {cos}2A = \dfrac{ \sqrt{3} }{2} \\

\rm \: {cos}2A = cos\bigg(\dfrac{\pi}{6} \bigg) \\

We know,

\red{\boxed{ \rm{ \:cosx = cosy \: \rm\implies \:x = 2n\pi \pm \: y\:  \:  \: \forall \: \:  \:  n \in \: Z}}}\\

So, using this result, we get

\rm \: 2x = 2n\pi \pm \: \dfrac{\pi}{6} \:  \:  \: \forall \: n \in \: Z\\

\rm\implies \:\rm \: x = n\pi \pm \: \dfrac{\pi}{12} \:  \:  \: \forall \: n \in \: Z\\

\rule{190pt}{2pt}

Formulae Used :-

\boxed{ \rm{ \:sin(A + B) \: sin(A - B) =  {sin}^{2}A -  {sin}^{2}B \: }} \\

\boxed{ \rm{ \:cos(A + B) \: cos(A - B) =  {cos}^{2}A -  {sin}^{2}B \: }} \\

\boxed{ \rm{ \:cos \frac{\pi}{6} =  \frac{ \sqrt{3} }{2} \: }} \\

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi  \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}

Answered by nihasrajgone2005
6

Given determinant is

\begin{gathered}\rm \: 2\left|\begin{array}{ll}\tt\sin (A+B) &amp;\tt \cos (A+B) \\ \tt\cos (A-B) &amp; \tt\sin (A-B)\end{array}\right|+\sqrt{3}=0 \\ \end{gathered} </p><p>2 </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p>  </p><p>sin(A+B)</p><p>cos(A−B)</p><p></p><p>  </p><p>cos(A+B)</p><p>sin(A−B)</p><p></p><p>  </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> + </p><p>3</p><p></p><p> =0</p><p>

can be rewritten as

\begin{gathered}\rm \: 2\left|\begin{array}{ll}\tt\sin (A+B) &amp;\tt \cos (A+B) \\ \tt\cos (A-B) &amp; \tt\sin (A-B)\end{array}\right|= - \sqrt{3} \\ \end{gathered} </p><p>2 </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p>  </p><p>sin(A+B)</p><p>cos(A−B)</p><p></p><p>  </p><p>cos(A+B)</p><p>sin(A−B)</p><p></p><p>  </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> =− </p><p>3

\begin{gathered}\rm \: \left|\begin{array}{ll}\tt\sin (A+B) &amp;\tt \cos (A+B) \\ \tt\cos (A-B) &amp; \tt\sin (A-B)\end{array}\right|= -\dfrac{ \sqrt{3} }{2} \\ \end{gathered} </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p>  </p><p>sin(A+B)</p><p>cos(A−B)</p><p></p><p>  </p><p>cos(A+B)</p><p>sin(A−B)</p><p></p><p>  </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> =− </p><p>2</p><p>3

\begin{gathered}\rm \: sin(A + B)sin(A - B) - cos(A + B)cos(A - B)= -\dfrac{ \sqrt{3} }{2} \\ \end{gathered} </p><p>sin(A+B)sin(A−B)−cos(A+B)cos(A−B)=− </p><p>2</p><p>3

\begin{gathered}\rm \:({sin}^{2}A- {sin}^{2}B)-( {cos}^{2}A-{sin}^{2}B) = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered} </p><p>(sin </p><p>2</p><p> A−sin </p><p>2</p><p> B)−(cos </p><p>2</p><p> A−sin </p><p>2</p><p> B)=− </p><p>2</p><p>3

\begin{gathered}\rm \:{sin}^{2}A- {sin}^{2}B-{cos}^{2}A + {sin}^{2}B = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered} </p><p>sin </p><p>2</p><p> A−sin </p><p>2</p><p> B−cos </p><p>2</p><p> A+sin </p><p>2</p><p> B=− </p><p>2</p><p>3

\begin{gathered}\rm \:{sin}^{2}A-{cos}^{2}A = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered} </p><p>sin </p><p>2</p><p> A−cos </p><p>2</p><p> A=− </p><p>2</p><p>3

\begin{gathered}\rm \: - ( - {sin}^{2}A + {cos}^{2}A) = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered} </p><p>−(−sin </p><p>2</p><p> A+cos </p><p>2</p><p> A)=− </p><p>2</p><p>3

\begin{gathered}\rm \: {cos}^{2}A - {sin}^{2}A = \dfrac{ \sqrt{3} }{2} \\ \end{gathered} </p><p>cos </p><p>2</p><p> A−sin </p><p>2</p><p> A= </p><p>2</p><p>3

\begin{gathered}\rm \: {cos}2A = \dfrac{ \sqrt{3} }{2} \\ \end{gathered} </p><p>cos2A= </p><p>2</p><p>3

\begin{gathered}\rm \: {cos}2A = cos\bigg(\dfrac{\pi}{6} \bigg) \\ \end{gathered} </p><p>cos2A=cos( </p><p>6</p><p>π</p><p></p><p> )</p><p>

We know,

\begin{gathered}\red{\boxed{ \rm{ \:cosx = cosy \: \rm\implies \:x = 2n\pi \pm \: y\: \: \: \forall \: \: \: n \in \: Z}}}\\\end{gathered} </p><p>cosx=cosy⟹x=2nπ±y∀n∈Z

So, using this result, we get

\begin{gathered}\rm \: 2x = 2n\pi \pm \: \dfrac{\pi}{6} \: \: \: \forall \: n \in \: Z\\\end{gathered} </p><p>2x=2nπ± </p><p>6</p><p>π</p><p></p><p> ∀n∈Z

\begin{gathered}\rm\implies \:\rm \: x = n\pi \pm \: \dfrac{\pi}{12} \: \: \: \forall \: n \in \: Z\\\end{gathered} </p><p>⟹x=nπ± </p><p>12</p><p>π</p><p></p><p> ∀n∈Z</p><p></p><p> </p><p></p><p>\rule{190pt}{2pt}

Formulae Used :-

\begin{gathered}\boxed{ \rm{ \:sin(A + B) \: sin(A - B) = {sin}^{2}A - {sin}^{2}B \: }} \\ \end{gathered} </p><p>sin(A+B)sin(A−B)=sin </p><p>2</p><p> A−sin </p><p>2</p><p> B

\begin{gathered}\boxed{ \rm{ \:cos(A + B) \: cos(A - B) = {cos}^{2}A - {sin}^{2}B \: }} \\ \end{gathered} </p><p>cos(A+B)cos(A−B)=cos </p><p>2</p><p> A−sin </p><p>2</p><p> B</p><p>

\begin{gathered}\boxed{ \rm{ \:cos \frac{\pi}{6} = \frac{ \sqrt{3} }{2} \: }} \\ \end{gathered} </p><p>cos </p><p>6</p><p>π</p><p></p><p> = </p><p>2</p><p>3

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq &amp; \bf Solution \\ \frac{\qquad \qquad}{} &amp; \frac{\qquad \qquad}{} \\ \sf sinx = 0 &amp; \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 &amp; \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 &amp; \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny &amp; \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy &amp; \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany &amp; \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}\end{gathered} </p><p>T−eq</p><p></p><p> </p><p>sinx=0</p><p>cosx=0</p><p>tanx=0</p><p>sinx=siny</p><p>cosx=cosy</p><p>tanx=tany</p><p></p><p>  </p><p>Solution</p><p></p><p> </p><p>x=nπ∀n∈Z</p><p>x=(2n+1) </p><p>2</p><p>π</p><p></p><p> ∀n∈Z</p><p>x=nπ∀n∈Z</p><p>x=nπ+(−1) </p><p>n</p><p> y∀n∈Z</p><p>x=2nπ±y∀n∈Z</p><p>x=nπ+y∀n∈Z

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