Question

please anyone give me the answer of this question,quickly.​

Answer 1

$\bullet\ \; \sf \dfrac{2.tan^260}{1-tan^230}+(sec^245-cot^245)+(sin^230+sin^260)$

$\to \sf \dfrac{2(\sqrt{3})^2}{1-\left(\frac{1}{\sqrt{3}}\right)^2}+\left( (\sqrt{2})^2-(1)^2 \right)+\left( \left(\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2 \right)$

$\to \sf \dfrac{2(3)}{1-\frac{1}{3}}+(2-1)+\left( \dfrac{1}{4}+\dfrac{3}{4} \right)$

$\to \sf \dfrac{2(3)}{\frac{2}{3}}+1+\dfrac{4}{4}$

$\to \sf 9+1+1\\\\\to\ \; \; \sf \orange{11}\ \; \bigstar$

Trigonometric Table :

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

Answer 2

S O L U T I O N :

Firstly, we all knows values which given in question ;

• tan 60° = √3
• tan 30° = 1/√3
• sec 45° = √2
• cot 45° = 1
• sin 30° = 1/2
• sin 60° = √3/2

$\underline{\underline{\tt{Accroding\:to\:the\:question\::}}}$

$\mapsto\tt{\dfrac{2tan^{2} \:60\degree }{1-tan^{2} \:30\degree } + (sec^{2} \:45\degree - cot^{2} \:45\degree ) + (sin^{2} \:30\degree + sin^{2} \: 60\degree )}$

$\mapsto\tt{\dfrac{2(\sqrt{3})^{2} }{1-(1/\sqrt{3} )^{2}} + [(\sqrt{2})^{2} - (1)^{2} ] + \bigg[\bigg(\dfrac{1}{2} \bigg)^{2} + \bigg(\dfrac{\sqrt{3} }{2}\bigg)^{2} \bigg]}$

$\mapsto\tt{\dfrac{2\times 3 }{1- \dfrac{1}{3}} + [2- 1 ] + \bigg[\dfrac{1}{4} + \dfrac{3}{4}\bigg]}$

$\mapsto\tt{\dfrac{6 }{\dfrac{3-1}{3}} + 1+ \bigg[\dfrac{1+3}{4} \bigg]}$

$\mapsto\tt{\dfrac{6 }{\dfrac{2}{3}} + 1+ \cancel{\bigg[\dfrac{4}{4} \bigg]}}$

$\mapsto\tt{\dfrac{6\times 3}{2} +1 + 1}$

$\mapsto\tt{\cancel{\dfrac{18}{2}} +1 + 1}$

$\mapsto\tt{9 + 2}$

$\mapsto\bf{11}$

Thus,

The value will be 11 .