Question

tan theta -sin square theta=cos square theta then show that sin square theta= 1/2

Answer 1

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

$\rm \: tan\theta - {sin}^{2}\theta = {cos}^{2}\theta$

$\large\underline{\sf{To\:show - }}$

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

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

Given that,

$\rm \: tan\theta - {sin}^{2}\theta = {cos}^{2}\theta$

$\rm \: tan\theta = {sin}^{2}\theta + {cos}^{2}\theta$

$\rm\implies \:tan\theta = 1$

$\rm\implies \:tan\theta = tan45 \degree$

$\rm\implies \:\theta = 45 \degree$

Now, Consider

$\rm \: {sin}^{2}\theta$

$\rm \: = \: {sin}^{2}45 \degree$

$\rm \: = \: {\bigg(\dfrac{1}{ \sqrt{2} } \bigg) }^{2}$

$\rm \: = \: \dfrac{1}{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \: {sin}^{2}\theta \: = \: \frac{1}{2} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$