Prove that:
[tex] {\rightarrow{ \large{ \rm{ \frac{ \sec \: \theta}{ \sec \: \theta \: + \: \tan \: \theta} \: + \: \frac{ \tan \: \theta }{ \sec \: \theta \: - \: \tan \: \theta } = \frac{ \sec \: \theta }{ \sec \: \theta \: - \: \tan \: \theta} \: - \: \frac{ \tan \: \theta }{ \sec \: \theta \: + \: \tan \: \theta } \: = \: \frac{1 + { \sin }^{2} \: \theta}{1 - { \sin }^{2} \: \theta} }}}}[/tex]
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[tex] {\rightarrow{ \large{ \rm{ \frac{ \sec \: \theta}{ \sec \: \theta \: + \: \tan \: \theta} \: + \: \frac{ \tan \: \theta }{ \sec \: \theta \: - \: \tan \: \theta } = \frac{ \sec \: \theta }{ \sec \: \theta \: - \: \tan \: \theta} \: - \: \frac{ \tan \: \theta }{ \sec \: \theta \: + \: \tan \: \theta } \: = \: \frac{1 + { \sin }^{2} \: \theta}{1 - { \sin }^{2} \: \theta} }}}}[/tex]
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Answers & Comments
Verified answer
Answer:
First Member:
[tex] { \large{ \dashrightarrow{ \rm{ \frac{ \sec \: \theta }{ \sec \: \theta \: + \: \tan \: \theta } \: + \: \frac{ \tan \: \theta }{ \sec \: \theta \: - \: \tan \: \theta } }}}}[/tex]
Taking L.C.M.
[tex]{ \large{ \longrightarrow{ \rm{ \frac{ \sec \: \theta \: ( \sec \: \theta \: - \tan \: \theta ) \: + \: \tan \: \theta \: ( \sec \: \theta \: + \tan \: \theta)}{( \sec \: \theta \: - \: \tan \: \theta )( \sec \: \theta \: - \: \tan \: \theta) } }}}}[/tex]
[tex]{ \large{\longrightarrow { \rm{ \frac{ { \sec}^{2} \: \theta \: - \: \sec \theta \: \tan \theta \: + \: \sec \theta \: \tan \theta \: + \: { \tan}^{2} \: \theta }{ { \sec }^{2} \: \theta \: - \: { \tan }^{2} \: \theta } }}}}[/tex]
[tex]{ \large{ \longrightarrow{ \rm{ \frac{ { \sec }^{2} \: \theta \: + \: { \tan }^{2} \: \theta}{ { \sec }^{2} \: \theta \: - \: { \tan }^{2} \: \theta} \: = \: \frac{ { \sec }^{2} \: \theta \: + \: { \tan }^{2} \: \theta}{1} }}}}[/tex]
[tex] { \large{ \longrightarrow{ \rm{ \frac{1}{ { \ \cos }^{2} \: \theta} \: + \: \frac{ { \sin }^{2} \: \theta}{ { \cos}^{2} \: \theta} \: = \: \frac{1 + { \sin}^{2} \: \theta }{ { \cos}^{2} \: \theta} \: = \: \frac{1 + { \sin }^{2} \: \theta}{1 - { \sin }^{2} \: \theta } \: \: \: \: \: \: ---(1)}}}}[/tex]
[tex]{ \large{ \longrightarrow{ \sf{Third \: \: Member }}}}[/tex]
Again Second Member:
[tex]{ \large{ \dashrightarrow{ \rm{ \frac{ \sec \: \theta}{ \sec \: \theta \: - \: \tan \: \theta } \: - \: \frac{ \tan \: \theta}{ \sec \: \theta \: + \: \tan \: \theta } }}}}[/tex]
Taking L.C.M.
[tex]{ \large{ \longrightarrow{ \rm{ \frac{ \sec \: \theta \: ( \sec \: \theta \: + \: \tan \: \theta) \: - \: \tan \: \theta \: ( \sec \: \theta \: - \: \tan \: \theta) }{ (\sec \: \theta \: - \: \tan \: \theta)( \sec \: \theta \: + \: \tan \: \theta) } }}}}[/tex]
[tex]{ \large{ \longrightarrow{ \rm{ \frac{ { \sec }^{2} \: \theta \: + \: \sec \theta \: \tan \theta \: - \: \sec \theta \: \tan \theta \: + \: { \tan }^{2} \theta }{ { \sec }^{2} \: \theta \: - \: { \tan }^{2} \: \theta } }}}}[/tex]
[tex]{ \large{ \longrightarrow{\rm{ \frac{ { \sec}^{2} \: \theta \: + \: { \tan }^{2} \: \theta }{ { \sec }^{2} \: \theta \: - \: { \tan }^{2} \: \theta } \: = \: { \sec }^{2} \: \theta \: + \: { \tan}^{2} \: \theta }}}}[/tex]
[tex]{ \large{ \longrightarrow{ \rm{ \frac{1}{ { \cos}^{2} \: \theta } \: + \: \frac{ { \sin}^{2} \: \theta}{ { \cos }^{2} \: \theta } \: = \: \frac{1 + { \sin}^{2} \: \theta}{ { \cos}^{2} \: \theta } \: = \: \frac{1 + { \sin }^{2} \: \theta}{1 - { \sin}^{2} \: \theta } \: \: \: \: \: \: ---(2) }}}}[/tex]
[tex]{ \large{ \longrightarrow{ \sf{Third \: \: Member}}}}[/tex]
From (1) and (2), we get:
[tex]{ \large{ \rm{ \boxed{ \frac{ \sec \: \theta}{ \sec \: \theta \: + \: \tan \: \theta } \: + \: \frac{ \tan \: \theta }{ \sec \: \theta \: - \: \tan \: \theta } \: = \frac{ \sec \: \theta}{ \sec \: \theta \: - \: \tan \: \theta } \: - \: \frac{ \tan \: \theta }{ \sec \: \theta \: + \: \tan \: \theta} \: = { \green{ \boxed{ \rm{ \frac{1 + { \sin }^{2} \: \theta}{1 - { \sin}^{2} \: \theta } }}}} }}}}[/tex]
Hence Proved!!
[tex]\large\underline{\sf{Solution-}}[/tex]
Consider LHS
[tex]\sf \: \dfrac{ \sec \: \theta}{ \sec \: \theta \: + \: \tan \: \theta} \: + \: \dfrac{ \tan \: \theta }{ \sec \: \theta \: - \: \tan \: \theta } \\ \\ [/tex]
can be rewritten as on taking LCM
[tex]\sf \: = \: \dfrac{ \sec \: \theta(\sec \: \theta - \tan \: \theta) + \tan \: \theta(\sec \: \theta + \tan \: \theta)}{ (\sec \: \theta \: + \: \tan \: \theta)(\sec \: \theta - \tan \: \theta)} \: \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ \sec \:^{2} \theta - \sec \: \theta\tan \: \theta + \tan \: \theta\sec \: \theta + \tan \:^{2} \theta}{ (\sec \: \theta \: + \: \tan \: \theta)(\sec \: \theta - \tan \: \theta)} \: \\ \\ [/tex]
can be re-arranged as
[tex]\sf \: = \: \dfrac{( \sec \:^{2} \theta + \sec \: \theta\tan \: \theta )+( - \tan \: \theta\sec \: \theta + \tan \:^{2} \theta)}{ (\sec \: \theta \: + \: \tan \: \theta)(\sec \: \theta - \tan \: \theta)} \: \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ \sec \: \theta(\sec \: \theta + \tan \: \theta) - \tan \: \theta(\sec \: \theta - \tan \: \theta)}{ (\sec \: \theta \: + \: \tan \: \theta)(\sec \: \theta - \tan \: \theta)} \: \\ \\ [/tex]
[tex]\bf \: = \: \dfrac{ \sec \: \theta }{ \sec \: \theta \: - \: \tan \: \theta} \: - \: \dfrac{ \tan \: \theta }{ \sec \: \theta \: + \: \tan \: \theta } \: \\ \\ [/tex]
Hence, Proved the first part.
Now, for the second part,
Consider LHS
[tex]\sf \: \dfrac{ \sec \: \theta}{ \sec \: \theta \: + \: \tan \: \theta} \: + \: \dfrac{ \tan \: \theta }{ \sec \: \theta \: - \: \tan \: \theta } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ \sec \: \theta(\sec \: \theta - \tan \: \theta) + \tan \: \theta(\sec \: \theta + \tan \: \theta)}{ (\sec \: \theta \: + \: \tan \: \theta)(\sec \: \theta - \tan \: \theta)} \: \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{ \sec \:^{2} \theta - \sec \: \theta\tan \: \theta + \tan \: \theta\sec \: \theta + \tan \:^{2} \theta}{ \sec \:^{2} \theta \: - \: \tan \:^{2} \theta} \: \\ \\ [/tex]
[tex]\sf \: = \: \sec \:^{2} \theta + \tan \: ^{2} \theta \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{\cos \:^{2}\theta} + \dfrac{ { \sin} \: ^{2}\theta }{ { \cos} \: ^{2}\theta} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1 + { \sin} \: ^{2}\theta }{ { \cos} \: ^{2}\theta} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1 + { \sin} \: ^{2}\theta }{1 - { \sin} \: ^{2}\theta} \\ \\ [/tex]
Hence,
[tex]\sf \:\dfrac{ \sec \: \theta}{ \sec \: \theta \: + \: \tan \: \theta} \: + \: \dfrac{ \tan \: \theta }{ \sec \: \theta \: - \: \tan \: \theta } = \: \dfrac{1 + { \sin} \: ^{2}\theta }{1 - { \sin} \: ^{2}\theta} \\ \\ [/tex]
Hence, Proved
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \mathfrak{Additional\:Information}}}[/tex]
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{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}[/tex]