Question

If 7sin^2thita+3cos^2thita=4 prove that tanthita=1/root3

Answer 1

Step-by-step explanation:

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Answer 2

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{7 {sin}^{2}\theta + 3 {cos}^{2}\theta = 4} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Prove - \begin{cases} &\sf{tan\theta = \dfrac{1}{ \sqrt{3} } }\end{cases}\end{gathered}\end{gathered}$

Identities Used :-

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

$\boxed{ \red{ \bf \: sin30 \degree = \dfrac{1}{2}}}$

$\boxed{ \red{ \bf \: tan30 \degree = \dfrac{1}{ \sqrt{3} }}}$

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

Given that

$\rm :\longmapsto\:7 {sin}^{2} \theta + 3 {cos}^{2} \theta = 4$

$\rm :\longmapsto\:7 {sin}^{2} \theta + 3 (1 - {sin}^{2} \theta )= 4$

$\rm :\longmapsto\:7 {sin}^{2} \theta + 3 - 3{sin}^{2} \theta= 4$

$\rm :\longmapsto\:4 {sin}^{2} \theta = 1$

$\sf :\longmapsto\: {sin}^{2} \theta = \dfrac{1}{4}$

$\rm :\implies\:sin\theta = \dfrac{1}{2}$

$\rm :\longmapsto\:sin\theta = sin30 \degree$

$\rm :\implies\:\theta = 30\degree$

Now

Consider,

$\rm :\longmapsto\:tan\theta$

$\rm \:  \:  =  \:  \: tan \: 30\degree$

$\rm \:  \:  =  \:  \: \dfrac{1}{ \sqrt{3} }$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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

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

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