Question

3 - 4 sin square theta upon cos square theta equals to 3 minus 10 squared theta prove that ​

Answer 1

$\huge{\pink{\underline{\underline{QUESTION-}}}}$

$\bf{\purple{\implies\:\frac{3}{ \ { \cos}^{2} \theta } - \frac{4 \: { \sin }^{2} \theta }{ { \cos }^{2} \theta}= 3 - {tan}^{2} \theta }}$

$\huge{\pink{\underline{\underline{SOLUTION-}}}}$

$\huge{\underline{\underline{L.H.S.-}}}$

$\mathrm{ = \frac{3}{ \ { \cos}^{2} \theta } - \frac{4 \: { \sin }^{2} \theta }{ { \cos }^{2} \theta}}$

$\mathrm{= 3 \: {sec}^{2} \theta - 4 \: {tan}^{2} \theta}$

$\mathsf{\red{ (Since,\:\implies \frac{1}{ { \cos}^{2} \theta} = { \sec}^{2} \theta\:and\:\frac{\: { \sin }^{2} \theta }{ { \cos }^{2} \theta}={\tan}^{2}\theta)}}$

$\mathrm {= 3 \: {sec}^{2} \theta - 3 \: {tan}^{2} \theta - { \tan }^{2} \theta}$

$\mathrm{= 3({sec}^{2} \theta - {tan}^{2} \theta) - { \tan }^{2} \theta}$

$\mathrm{= 3 \times 1 - {tan}^{2} \theta}$

$\mathrm{= 3 - {tan}^{2} \theta}$

$\mathcal{\implies\:L.H.S.=R.H.S.}$

$\huge{\texttt{Hence\:\:Proved}}$

Answer 2

$\huge{ \purple{ \underline{ \underline{QUESTION-}}}}$

$\large \pink{= > \frac{3}{ {cos}^{2}\theta } - \frac{4 { \sin }^{2}\theta }{ {cos }^{2} \theta} = \: 3 \: - {tan\theta \: }^{2} }$

$\huge{ \purple{ \underline{ \underline{SOLUTION-}}}}$

$\huge\underline\red{R.H.S\:-}$

$\large\pink{=> 3\: - \frac{{sin}^{2}\theta}{{Cos}^{2}\theta}}$

$\large\pink{=> \frac{3{cos}^{2}\theta\: -{sin}^{2}\theta}{{cos}^2}\theta}$

$\large\pink{=>\frac{3(1-{sin}^{2}\theta)-{sin}^{2}\theta}{{cos}^{2}}}$

$\large\pink{=> \frac{3-4{sin}^{2}}{{cos}^{2}}}$

$\large\purple{=>\frac{3}{{cos}^{2}}-\frac{4{sin}^{2}}{{cos}^{2}}}$

$\huge{L.H.S= R.H.S}$

$\huge\red{PROVED}$