Question

solve this...........​

Answer 1

Answer:

xvshhsgsvsbdbs sbzv a kada problem ie a trgsbshs

Answer 2

${ \blue{ \underline{ \underline{{ \pink{{ \huge{ \mathfrak{ \: \: S}}} \bold{OLUTION - \: \: \: }}}}}}}$

${ \tt{ \theta - \phi = \frac{\pi}{2} }} \\ \\ \implies{ \tt{ - \phi = \frac{\pi}{2} - \theta }} \\ \\ \implies{ \tt{ \phi = - ( \frac{\pi}{2} -\theta) }} \\ \\ \boxed{ \implies{ \tt{ \phi = - \frac{\pi}{2} - \theta}}}$

To prove :-

${ \tt{\left[ \begin{array}{c c } \sf { cos}^{2}\theta & \sf cos \theta \: sin \theta\\\\ \sf \: cos\theta \: sin\theta& \sf \: {sin}^{2} \theta \end{array} \right] \left[ \begin{array}{c c } \sf {cos}^{2} \phi & \sf cos \phi \: sin \phi \\\\ \sf \: cos \phi \: sin \phi& \sf \: {sin}^{2} \phi \end{array} \right] = 0}}$

$\implies{ \tt{\left[ \begin{array}{c c } \sf { cos}^{2}\theta & \sf cos \theta \: sin \theta\\\\ \sf \: cos\theta \: sin\theta& \sf \: {sin}^{2} \theta \end{array} \right] \left[ \begin{array}{c c } \sf {cos}^{2} ( \frac{\pi}{2} - \theta) & \sf cos( \frac{\pi}{2} - \theta) \: sin ( \frac{\pi}{2} - \theta) \\\\ \sf \: cos ( \frac{\pi}{2} - \theta)\: sin ( \frac{\pi}{2} - \theta)& \sf \: {sin}^{2} ( \frac{\pi}{2} - \theta)\end{array} \right] = 0}} \\ \\ \\ \implies{ \tt{{ \tt{\left[ \begin{array}{c c } \sf { cos}^{2}\theta & \sf cos \theta \: sin \theta\\\\ \sf \: cos\theta \: sin\theta& \sf \: {sin}^{2} \theta \end{array} \right] \left[ \begin{array}{c c } \sf {sin}^{2} \theta& \sf sin \theta\: cos\theta\\\\ \sf \: sin \theta\: cos\theta& \sf \: {cos}^{2} \theta \end{array} \right] = 0}}}} \\ \\ \\ \implies{ \tt{{ \tt{\left[ \begin{array}{c c } \sf { cos}^{2}\theta. {sin}^{2} \theta - { cos}^{2}\theta. {sin}^{2} \theta && \sf { cos}^{2}\theta. {sin}^{2} \theta - { cos}^{2}\theta. {sin}^{2} \theta\\\\ \sf { cos}^{2}\theta. {sin}^{2} \theta - { cos}^{2}\theta. {sin}^{2} \theta && \sf { cos}^{2}\theta. {sin}^{2} \theta - { cos}^{2}\theta. {sin}^{2} \theta \end{array} \right] = 0}}}} \\ \\ \\ \implies{ \tt{\left[ \begin{array}{c c } \sf 0& \sf 0 \\\\ \sf \: 0& \sf \: 0 \end{array} \right] = 0}}$

(proved)

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@MissSolitary✌️