Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given determinant is

[tex]\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 \\ [/tex]

can be rewritten as

[tex]\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} \\ [/tex]

[tex]\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} \\ [/tex]

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

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

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

[tex]\rm \:{sin}^{2}A-{cos}^{2}A = -\dfrac{ \sqrt{3} }{2} \\ [/tex]

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

[tex]\rm \: {cos}^{2}A - {sin}^{2}A = \dfrac{ \sqrt{3} }{2} \\ [/tex]

[tex]\rm \: {cos}2A = \dfrac{ \sqrt{3} }{2} \\ [/tex]

[tex]\rm \: {cos}2A = cos\bigg(\dfrac{\pi}{6} \bigg) \\ [/tex]

We know,

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

So, using this result, we get

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

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

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used :-

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

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

[tex]\boxed{ \rm{ \:cos \frac{\pi}{6} = \frac{ \sqrt{3} }{2} \: }} \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\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}[/tex]

Given determinant is

[tex]\begin{gathered}\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 \\ \end{gathered}

2

∣

∣

∣

∣

∣

sin(A+B)

cos(A−B)

cos(A+B)

sin(A−B)

∣

∣

∣

∣

∣

+

3

=0

[/tex]

can be rewritten as

[tex]\begin{gathered}\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} \\ \end{gathered}

2

∣

∣

∣

∣

∣

sin(A+B)

cos(A−B)

cos(A+B)

sin(A−B)

∣

∣

∣

∣

∣

=−

3[/tex]

[tex]\begin{gathered}\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} \\ \end{gathered}

∣

∣

∣

∣

∣

sin(A+B)

cos(A−B)

cos(A+B)

sin(A−B)

∣

∣

∣

∣

∣

=−

2

3[/tex]

[tex]\begin{gathered}\rm \: sin(A + B)sin(A - B) - cos(A + B)cos(A - B)= -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}

sin(A+B)sin(A−B)−cos(A+B)cos(A−B)=−

2

3[/tex]

[tex]\begin{gathered}\rm \:({sin}^{2}A- {sin}^{2}B)-( {cos}^{2}A-{sin}^{2}B) = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}

(sin

2

A−sin

2

B)−(cos

2

A−sin

2

B)=−

2

3[/tex]

[tex]\begin{gathered}\rm \:{sin}^{2}A- {sin}^{2}B-{cos}^{2}A + {sin}^{2}B = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}

sin

2

A−sin

2

B−cos

2

A+sin

2

B=−

2

3[/tex]

[tex]\begin{gathered}\rm \:{sin}^{2}A-{cos}^{2}A = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}

sin

2

A−cos

2

A=−

2

3[/tex]

[tex]\begin{gathered}\rm \: - ( - {sin}^{2}A + {cos}^{2}A) = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}

−(−sin

2

A+cos

2

A)=−

2

3[/tex]

[tex]\begin{gathered}\rm \: {cos}^{2}A - {sin}^{2}A = \dfrac{ \sqrt{3} }{2} \\ \end{gathered}

cos

2

A−sin

2

A=

2

3[/tex]

[tex]\begin{gathered}\rm \: {cos}2A = \dfrac{ \sqrt{3} }{2} \\ \end{gathered}

cos2A=

2

3[/tex]

[tex]\begin{gathered}\rm \: {cos}2A = cos\bigg(\dfrac{\pi}{6} \bigg) \\ \end{gathered}

cos2A=cos(

6

π

)

[/tex]

We know,

[tex]\begin{gathered}\red{\boxed{ \rm{ \:cosx = cosy \: \rm\implies \:x = 2n\pi \pm \: y\: \: \: \forall \: \: \: n \in \: Z}}}\\\end{gathered}

cosx=cosy⟹x=2nπ±y∀n∈Z[/tex]

So, using this result, we get

[tex]\begin{gathered}\rm \: 2x = 2n\pi \pm \: \dfrac{\pi}{6} \: \: \: \forall \: n \in \: Z\\\end{gathered}

2x=2nπ±

6

π

∀n∈Z[/tex]

[tex]\begin{gathered}\rm\implies \:\rm \: x = n\pi \pm \: \dfrac{\pi}{12} \: \: \: \forall \: n \in \: Z\\\end{gathered}

⟹x=nπ±

12

π

∀n∈Z

\rule{190pt}{2pt}[/tex]

Formulae Used :-

[tex]\begin{gathered}\boxed{ \rm{ \:sin(A + B) \: sin(A - B) = {sin}^{2}A - {sin}^{2}B \: }} \\ \end{gathered}

sin(A+B)sin(A−B)=sin

2

A−sin

2

B[/tex]

[tex]\begin{gathered}\boxed{ \rm{ \:cos(A + B) \: cos(A - B) = {cos}^{2}A - {sin}^{2}B \: }} \\ \end{gathered}

cos(A+B)cos(A−B)=cos

2

A−sin

2

B

[/tex]

[tex]\begin{gathered}\boxed{ \rm{ \:cos \frac{\pi}{6} = \frac{ \sqrt{3} }{2} \: }} \\ \end{gathered}

cos

6

π

=

2

3[/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\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}\end{gathered}

T−eq

sinx=0

cosx=0

tanx=0

sinx=siny

cosx=cosy

tanx=tany

Solution

x=nπ∀n∈Z

x=(2n+1)

2

π

∀n∈Z

x=nπ∀n∈Z

x=nπ+(−1)

n

y∀n∈Z

x=2nπ±y∀n∈Z

x=nπ+y∀n∈Z[/tex]