Question

$if \: 7sin {}^{2} theta + 5cos {}^{2}theta = 4 \: then \: evaluate \: sectheta + \csc \: theta$
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Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{7 {sin}^{2}\theta + 5 {cos}^{2}\theta = 4} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{sec\theta + cosec\theta}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \sf \: {sin}^{2}x + {cos}^{2}x = 1}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:7 {sin}^{2}\theta + 5 {cos}^{2}\theta = 4$

$\rm :\longmapsto\:7 {sin}^{2}\theta + 5(1 - {sin}^{2}\theta) = 4$

$\rm :\longmapsto\:7 {sin}^{2}\theta + 5 - 5{sin}^{2}\theta = 4$

$\rm :\longmapsto\:2 {sin}^{2}\theta = 4 - 5$

$\rm :\longmapsto\:2 {sin}^{2}\theta = - \: 1$

As squared number can never be negative,

So,

$\bf\implies \:There \: is \: no \: real \: value \: of \: \theta$

Additional Information :-

$\boxed{ \sf \: {sec}^{2}x - {tan}^{2}x = 1}$

$\boxed{ \sf \: {cosec}^{2}x - {cot}^{2}x = 1}$

$\boxed{ \sf \: cotx = \dfrac{1}{tanx}}$

$\boxed{ \sf \: cosecx = \dfrac{1}{sinx}}$

$\boxed{ \sf \: secx = \dfrac{1}{cosx}}$

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