Question

If 0° < α < 90° , then find the value of the given identity :
$\rm \dfrac{ \tan( \alpha ) - \sec( \alpha ) - 1 }{ \tan( \alpha ) + \sec( \alpha ) + 1 }$

- SSC CGL Tier II. Need Urgently​

Answer 1

AnswEr :

$\Longrightarrow\dfrac{ \tan( \alpha ) - \sec( \alpha ) - 1 }{ \tan( \alpha ) + \sec( \alpha ) + 1 }$

⠀⠀⠀⠀⋆ (sec² α - tan² α = 1)

$\Longrightarrow\dfrac{ \tan( \alpha ) - \sec( \alpha ) - ( { \sec}^{2}( \alpha ) - \tan^{2} ( \alpha )) }{ \tan( \alpha ) + \sec( \alpha ) + 1 }$

$\Longrightarrow\dfrac{ \tan( \alpha ) - \sec( \alpha ) - ( \sec( \alpha ) + \tan( \alpha ))( \sec( \alpha ) - \tan( \alpha ) ) }{ \tan( \alpha ) + \sec( \alpha ) + 1 }$

⠀⠀⠀⠀⋆ Taking Negative Common

$\Longrightarrow\dfrac{ - (\sec( \alpha ) - \tan( \alpha )) - ( \sec( \alpha ) + \tan( \alpha ))( \sec( \alpha ) - \tan( \alpha ) ) }{ \tan( \alpha ) + \sec( \alpha ) + 1 }$

⠀⠀⠀⠀⋆ Taking (sec α - tan α) Common

$\Longrightarrow\dfrac{(1 - ( \sec( \alpha ) + \tan( \alpha ))(\sec( \alpha) - \tan( \alpha)) }{ \tan( \alpha ) + \sec( \alpha ) + 1 }$

$\Longrightarrow\dfrac{(1 - \sec( \alpha ) - \tan( \alpha ))(\sec( \alpha) - \tan( \alpha ))}{ \tan( \alpha ) + \sec( \alpha ) + 1 }$

⠀⠀⠀⠀⋆ Taking Negative Common

$\Longrightarrow\dfrac{ - \cancel{(1 + \sec( \alpha ) + \tan( \alpha ))} (\sec( \alpha) - \tan( \alpha ))} {\cancel{ \tan( \alpha ) + \sec( \alpha ) + 1 }}$

$\Longrightarrow - ( \sec( \alpha ) - \tan( \alpha ) )$

$\Longrightarrow - \sec( \alpha ) + \tan( \alpha )$

$\Longrightarrow \tan( \alpha ) - \sec( \alpha )$

⠀⠀⠀⠀⋆ tan α = sin α / cos α

⠀⠀⠀⠀⋆ sec α = 1 / cos α

$\Longrightarrow \dfrac{ \sin( \alpha ) }{ \cos( \alpha ) } + \dfrac{1}{ \cos( \alpha ) }$

$\Longrightarrow \large \dfrac{ \sin( \alpha ) - 1 }{ \cos( \alpha ) }$

⠀

$\therefore \boxed{\dfrac{ \tan( \alpha ) - \sec( \alpha ) - 1 }{ \tan( \alpha ) + \sec( \alpha ) + 1 } = \dfrac{ \sin( \alpha ) - 1}{ \cos( \alpha ) } }$

Answer 2

$\underline{{\colorbox{yellow}{Question}}}$

$\rm \dfrac{ \tan( \alpha ) - \sec( \alpha ) - 1 }{ \tan( \alpha ) + \sec( \alpha ) + 1 }$

$\LARGE\underline{\underline{\sf \red{S}\blue{o}\green{l}\orange{u}\pink{t}\purple{i}\orange{o}\red{n}:}}$

Formula To be used ----

• Sec²A - tan²A = 1
• Sec A = 1/cos A
• Tan A = Sin A/Cos A

$\rm \dfrac{ \tan( \alpha ) - \sec( \alpha ) - 1 }{ \tan( \alpha ) + \sec( \alpha ) + 1 } \\ \\ putting \: value \: of \: 1 \: in \: numerator \\ \\ \rm \dfrac{ \tan( \alpha ) - \sec( \alpha ) - ( \sec^{2} \alpha - \tan^{2} \alpha) }{ \tan( \alpha ) + \sec( \alpha ) + 1 } \\ \\ using \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: now \\ \\ \rm \dfrac{( \tan( \alpha ) - \sec( \alpha )) - ((\sec( \alpha ) - \tan( \alpha )(\tan( \alpha ) + \sec( \alpha ))}{ \tan( \alpha ) + \sec( \alpha ) + 1 } \\ \\ taking \: ( - ve) \: from \: (\sec( \alpha ) - \tan( \alpha) \\ of \: numerator \: now \: and \: than \: \\ taking \: ( \tan( \alpha ) - \sec( \alpha )) \: also \: common \\ we \: get \implies \\ \\ \frac{ ( \tan( \alpha ) - \sec( \alpha )) \cancel{(1 + (( \tan( \alpha ) + \sec( \alpha )))}}{\cancel{(1 + (( \tan( \alpha ) + \sec( \alpha )))}} \\ \\ \\ \implies \: ( \tan( \alpha ) - \sec( \alpha )) \\ \\ \implies \: \frac{sin( \alpha)}{cos( \alpha)} - \frac{1}{cos( \alpha)} \\ \\ \implies \: \frac{sin( \alpha) - 1}{cos( \alpha)} \:$

$\color {red}\large\bold\star\underline\mathcal{Extra\:Brainly\:Knowledge:-}$

$\underline\textsf{Double Angle Identities}$

→ sin²θ = 2sinθcosθ

= $\frac{2 \tanθ}{1+ \tan^{2}θ}$

→ cos2θ = cos²θ- sin²θ

= 1 - 2sin²θ

= 2cos²θ- 1

= $\frac{1- \tan^{2}θ}{1+ \tan^{2}θ}$

→Tan2θ = $\frac{2 \tanθ}{1- \tan^{2}θ}$

$\mathcal{BE\:\:BRAINLY}$