Question

if cosec theta - sin theta and sec theta - cos theta = m prove that l²m²(l²+m²+3)=1

do this plz​

Answer 1

We have l = cosec θ - sin θ and m = sec θ - cos θ

$\therefore \mathrm{LHS} = \big( {l}^{2} + {m}^{2} + 3 \big)$

$: \longmapsto \rm{LHS= \bigg \lgroup \dfrac{1}{ \sin \theta} - \sin \theta \bigg \rgroup^{2} \bigg \lgroup \dfrac{1}{ \cos \theta} - \cos \theta \bigg \rgroup^{2} \left \{\begin{align} {} \bigg \lgroup \dfrac{1}{ \sin \theta} - \sin \theta \bigg \rgroup^{2} + \bigg \lgroup \dfrac{1}{ \cos \theta} - \cos \theta \bigg \rgroup^{2} + 3 \end{align} \right \}}$

$: \longmapsto \rm{LHS= \bigg \lgroup \dfrac{1- \sin^{2} \theta}{ \sin \theta} \bigg \rgroup^{2} \bigg \lgroup \dfrac{1 - \cos ^{2} \theta }{ \cos \theta} \bigg \rgroup^{2} \left \{\begin{align} {} \bigg \lgroup \dfrac{1- \sin^{2} \theta}{ \sin \theta} \bigg \rgroup^{2} + \bigg \lgroup \dfrac{1 - \cos ^{2} \theta }{ \cos \theta} \bigg \rgroup^{2} \end{align} \right \}}$

$: \longmapsto \rm{LHS= \bigg \lgroup \dfrac{ \cos^{2} \theta}{ \sin \theta} \bigg \rgroup^{2} \bigg \lgroup \dfrac{\sin ^{2} \theta }{ \cos \theta} \bigg \rgroup^{2} \left \{\begin{align} {} \bigg \lgroup \dfrac{ \cos^{2} \theta}{ \sin \theta} \bigg \rgroup^{2} + \bigg \lgroup \dfrac{\sin ^{2} \theta }{ \cos \theta} \bigg \rgroup^{2} + 3\end{align} \right \}}$

$: \longmapsto \rm{LHS= \dfrac{ \cancel{\cos^{4} \theta}}{ \cancel{\sin ^{2} \theta} } \times \dfrac{ \cancel{\sin ^{4} \theta}}{ \cancel{ \cos ^{2} \theta}} \left \{\begin{align} {} \dfrac{ \cos^{4} \theta}{ \sin ^{2} \theta} + \dfrac{ \sin ^{4} \theta}{ \cos ^{2} \theta} + 3 \end{align} \right \}}$

$: \longmapsto \rm{LHS= \cancel{ \cos^{2} \theta \times \sin ^{2} \theta} \left \{\begin{align} {} \dfrac{ \cos^{6} \theta + \sin ^{6} \theta + 3 \cos ^{2} \theta \sin ^{2} \theta }{ \cancel{\cos^{2} \theta \sin ^{2} \theta}} \end{align} \right \}}$

$: \longmapsto \rm{LHS= \cos^{6} \theta + \sin ^{6} \theta + 3 \cos ^{2} \theta \sin ^{2} \theta }$

$: \longmapsto \rm{LHS= \left \{\begin{align} (\cos^{2} \theta) ^{3} + ( \sin ^{2} \theta) ^{3} \end{align} \right \} + 3 \cos ^{2} \theta \sin ^{2} \theta }$

$\rm{ : \longmapsto LHS= \left \{\begin{align} \big(\cos^{2} \theta + \sin ^{2} \theta \big) ^{3} - 3 \cos ^{2} \theta \sin ^{2} \theta \big(\cos^{2} \theta + \sin ^{2} \theta \big) \end{align} \right \} + 3 \sin^{2} \theta \cos ^{2} \theta }$

$\rm{\big [ \because {a}^{3} + {b}^{3} = (a + b)^{3} - 3ab(a + b) \big] }$

$\rm{ : \longmapsto RHS= \left \{\begin{align} 1 - 3 \cos ^{2} \theta \sin ^{2} \theta \end{align} \right \} + 3 \sin^{2} \theta \cos ^{2} \theta } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big[ \because { \cos}^{2} \theta + { \sin}^{2} \theta = 1 \big]$

Answer 2

Calculation:

$\begin{aligned}   & \cos ec\theta -\sin </p><p>\theta =l \\ & \frac{1}{\sin \theta }-\sin \theta =l \\ & \frac{1-{{\sin }^{2}}\theta }{\sin \theta }=l \\ & \frac{{{\cos }^{2}}\theta }{\sin \theta }=l  \end{aligned}$

$Instruction: Simplify\sec \theta -\cos \theta =m$

Calculation:

$\begin{aligned}   & \sec \theta -\cos \theta =m \\ & \frac{1}{\cos \theta }-\cos \theta =m \\ & \frac{1-{{\cos }^{2}}\theta }{\cos \theta }=m \\ & \frac{{{\sin }^{2}}\theta }{\cos \theta }=m  \end{aligned}$

Instruction: Prove that,

${{l}^{2}}{{m}^{2}}\left( {{l}^{2}}+{{m}^{2}}+3 \right)=1$

Calculation:

$\begin{aligned}   & {{l}^{2}}{{m}^{2}}\left( {{l}^{2}}+{{m}^{2}}+3 \right)={{\left( \frac{{{\cos }^{2}}\theta }{\sin \theta } \right)}^{2}}{{\left( \frac{{{\sin }^{2}}\theta }{\cos \theta } \right)}^{2}}\left[ {{\left( \frac{{{\cos }^{2}}\theta }{\sin \theta } \right)}^{2}}+{{\left( \frac{{{\sin }^{2}}\theta }{\cos \theta } \right)}^{2}}+3 \right] \\ & =\left( \frac{{{\cos }^{4}}\theta }{{{\sin }^{2}}\theta } \right)\left( \frac{{{\sin }^{4}}\theta }{{{\cos }^{2}}\theta } \right)\left[ \frac{{{\cos }^{4}}\theta }{{{\sin }^{2}}\theta }+\frac{{{\sin }^{4}}\theta }{{{\cos }^{2}}\theta }+3 \right] \\ & ={{\cos }^{2}}\theta {{\sin }^{2}}\theta \left( \frac{{{\cos }^{4}}\theta }{{{\sin }^{2}}\theta } \right)+{{\cos }^{2}}\theta {{\sin }^{2}}\theta  \\ & \left( \frac{{{\sin }^{4}}\theta }{{{\cos }^{2}}\theta } \right)+3{{\cos }^{2}}\theta {{\sin }^{2}}\theta  \\ & ={{\cos }^{6}}\theta +{{\sin }^{6}}\theta +3{{\cos }^{2}}\theta {{\sin }^{2}}\theta   \end{aligned}$

Simplify further,

$\begin{aligned}   & {{l}^{2}}{{m}^{2}}\left( {{l}^{2}}+{{m}^{2}}+3 \right)={{\left( {{\cos }^{2}}\theta  \right)}^{3}}+{{\left( {{\sin }^{2}}\theta  \right)}^{3}}+3{{\cos }^{2}}\theta {{\sin }^{2}}\theta  \\ & ={{\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  \right)}^{3}}-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta  \\ & \left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  \right)+3{{\cos }^{2}}\theta {{\sin }^{2}}\theta  \\ &=1-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta +3{{\cos }^{2}}\theta {{\sin }^{2}}\theta  \\ & =1  \end{aligned}$

$It \: \: is \: \: verified \\: that, \: \: [Tex]</p><p>[tex]{{l}^{2}}{{m}^{2}}\left( {{l}^{2}}+{{m}^{2}}+3 \right)=1</p><p>.$