Question

prove this trignometric identity ​

Answer 1

$\huge {\underline{ \mathfrak{ \red s \blue o \green l \pink u \orange t \purple i \red t o \pink n : - }}}$

$\dfrac{1}{1 + \sin( \theta) } + \dfrac{1}{1 - \sin( \theta) }$

★ Taking LCM ,

$\longrightarrow \: \dfrac{1 + \sin( \theta) + 1 - \sin( \theta) }{[ 1 + \sin( \theta) ][ 1 - \sin( \theta) ] } \\ \\ \longrightarrow \: \frac{1 + 1 \cancel{ + \sin( \theta)} \cancel{ - \sin( \theta) } }{[ 1 + \sin( \theta) ][ 1 - \sin( \theta) ] }$

$\large {\underline {\boxed {\bigstar {\sf{(a + b)(a - b) = {a}^{2} - {b}^{2} }}}}}$

$\longrightarrow \: \dfrac{1 + 1}{1 - \sin {}^{2} ( \theta) }$

$\large {\underline {\boxed {\bigstar{ \sf{1 - \sin {}^{2} ( \theta) = \cos^{2}( \theta) }}}}}$

$\longrightarrow \: \dfrac{2}{ \cos {}^{2} ( \theta) } \\ \\ \longrightarrow \: 2 \times \frac{1}{ \cos {}^{2} ( \theta) }$

$\large {\underline {\boxed {\bigstar {\sf{ \: \dfrac{1}{ \cos {}^{2} ( \theta) } = { \sec}^{2}( \theta) }}}}}$

$\longrightarrow2 \: \sec {}^{2} ( \theta) \\ \\ \longrightarrow \: rhs$

$\dag \huge {\blue{\boxed {\green{\rm{HENCE \: \:PROVED}}}}}$

$\huge {\underline {\mathfrak{ \red s \blue o \green m \pink e \: \orange f \purple o \red rm \pink u \blue l \green a \red e : - }}}$

$\large \sf \: (1) \: \frac{ \sin( \theta) }{ \cos (\theta)} = \tan( \theta) \\ \\ \sf \large (2) \: \frac{ \cos( \theta) }{ \sin( \theta) } = \cot( \theta) \\ \\ \large \sf \: (3) \: \frac{1}{ \sin( \theta) } = \csc( \theta) \\ \\ \large \sf \: (4) \: 1 - \cos { }^{2} ( \theta) = \sin {}^{2} (\theta)$