Question

please refer to attachment and answer the question
no spam​

Answer 1

here is yur answer ...

hope it will help u

Answer 2

GIVEN :-

$\\ \bigstar \: \displaystyle \sf \dfrac{\tan ^{2} 60 ^{ \circ} + 4 \cos ^{2} 45^{ \circ} + 3 \sec ^{2} 30^{ \circ} + 5 \cos ^{2} 90^{ \circ} }{ \csc30^{ \circ} + \sec60^{ \circ} - \cot ^{2} 30^{ \circ} }$

TO FIND :-

$\\ \bigstar \: \displaystyle \sf \: value \: of \: \: \dfrac{\tan ^{2} 60 ^{ \circ} + 4 \cos ^{2} 45^{ \circ} + 3 \sec ^{2} 30^{ \circ} + 5 \cos ^{2} 90^{ \circ} }{ \csc30^{ \circ} + \sec60^{ \circ} - \cot ^{2} 30^{ \circ} }$

SOLUTION :-

$\\ \\ \begin{gathered}\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}\end{gathered}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\\ \\ : \implies \displaystyle \sf \dfrac{\tan ^{2} 60 ^{ \circ} + 4 \cos ^{2} 45^{ \circ} + 3 \sec ^{2} 30^{ \circ} + 5 \cos ^{2} 90^{ \circ} }{ \csc30^{ \circ} + \sec60^{ \circ} - \cot ^{2} 30^{ \circ} }$

$\displaystyle \sf : \implies \dfrac{( \sqrt{3}) ^{2} + 4 \times \bigg(\dfrac{1}{ \sqrt{2} } \bigg) ^{2} + 3 \times \bigg( \dfrac{2}{ \sqrt{3} } \bigg) ^{2} + 5 \times (0) ^{2} }{2 + 2 - (\sqrt{3}) ^{2} }$

$: \implies \displaystyle \sf \dfrac{3 + 4 \times \dfrac{1}{2} + 3 \times \dfrac{4}{3} + 5 \times 0 }{4 - (\sqrt{3}) ^{2} }$

$: \implies \displaystyle \sf \frac{3 + 2 \times 1 + 4}{4 - 3}$

$: \implies \displaystyle \sf \frac{3 + 2 + 4}{1}$

$\bigstar \:\underline{ \boxed{\displaystyle \sf \dfrac{\tan ^{2} 60 ^{ \circ} + 4 \cos ^{2} 45^{ \circ} + 3 \sec ^{2} 30^{ \circ} + 5 \cos ^{2} 90^{ \circ} }{ \csc30^{ \circ} + \sec60^{ \circ} - \cot ^{2} 30^{ \circ} } = 9 } }$

$\therefore \underline{\displaystyle \sf \: Required \: answer \: is \: 9 .\: }$