Question

answer this urgent fast​

Answer 1

$\huge\bf{\pink{\underline{\underline{\mathcal{AnSwer࿐}}}}}$

$\mathbb{GIVEN}$:-

$\longrightarrow$ $\bf SecA = √2$

$\longrightarrow$ $\bf Sec 45⁰ = √2$

$\longrightarrow$ hence, $\bf A = 45⁰$

$\mathbb{TO\:FIND}$:-

$\bf\dfrac{3Cos²A+5tan²A}{4tan²A-Sin²A}$

here,

$\longrightarrow$ CosA = Cos45⁰ = $\bf\dfrac{1}{√2}$

$\longrightarrow$ tanA = tan45⁰ = 1

$\longrightarrow$ SinA = Sin45⁰ = $\bf\dfrac{1}{2}$

$\mathbb{SOLUTION}$:-

$\longrightarrow$ $\bf\dfrac{3×\bigg(\dfrac{1}{√2}\bigg )^{2} +5(1)²}{4(1)²-\bigg(\dfrac{1}{√2}\bigg )^{2}}$

$\longrightarrow$ $\bf\dfrac{\dfrac{3}{2}+5}{4-\dfrac{1}{2}}$

$\longrightarrow$ $\bf\dfrac{\dfrac{3+10}{2}}{\dfrac{8-1}{2}}$

$\longrightarrow$ $\bf\dfrac{\dfrac{13}{2}}{\dfrac{7}{2}}$

$\longrightarrow$ $\bf\dfrac{13}{2}$ × $\bf\dfrac{2}{7}$

$\longrightarrow$ $\bf\dfrac{13}{7}$

_________________________

$\mathbb{ADDITIONAL\: INFORMATION}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

________________________

Hope it's helpful ツ

$\small\bf\pink{@itzsplendidcharm࿐}$

Answer 2

$\huge\sf \color{blue}\underline{\underline{Answer}} :$

$\sf\color{red} \ \ \ \ \ \ \ \ \ ⟹ \boxed{\frac{13}{7}}$

$\sf \bf \large \color{blue}\underline{ Given} :$

$\sf {Sec A =√2} \\ \sf » A = 45° \ \ \ \ \ \ \ \\ \sf{ (From \ Trigonometric \ table) }$

$\huge\sf \color{blue}\underline{\underline{Solution}} :$

$\\ \sf \large \frac{3Cos² A+5tan²A}{4tan²A-Sin²A} \\ \\ ⟹ \sf \large \frac{3Cos²45°+5tan²45°}{4tan²45°-Sin²45°}$

$\\ \sf \color{green}{ Now \ put \ the \ values \ from \ Trigonometric \ table }$$\\ \\ ⟹\sf{ \bf \frac{ \large 3×\frac{1²}{√2²} + 5× {1²}}{4×1²-\frac{1}{√2}²}}\\ \\ \large ⟹\sf{ \frac{\frac{3}{2}+5}{4- \frac{1}{2}}}\\ \\ \large ⟹{ \bf \sf{ \frac{\frac{3+10}{\cancel{2}}}{\frac{8-1}{\cancel{2}}}}} \\ \\ \large ⟹\bf \sf \frac{13}{7}$

$\color{green}━━━━━━━━━━━━━━━━━━━━━━━━━━$

Trigonometric Table

$\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}$

$\color{green}━━━━━━━━━━━━━━━━━━━━━━━━━━$