Question

sinθ−cosθ+1 / ​sinθ+cosθ−1 = 1/secθ−tanθ

prove that LHS=RHS
BY SOLVING RHS​

Answer 1

$\large\underline{\underline\blue{\sf We \: have : - }}$

sinθ−cosθ+1 / sinθ+cosθ−1 = 1/secθ−tanθ

$\large\underline{\underline\blue{\sf \: to \: find : - }}$

• prove that LHS = RHS

$\large\underline{\underline\green{\sf \: solution : - }}$

Let us deduce the expression from LHS towards RHS

Explanation:

Since we will apply the identity involving secθ and tanθ, first convert the LHS in terms of secθ and tanθ by dividing both numerator and denominator by cosθ.

thus,

$\sf \: LHS = \dfrac{ ( sinθ - cosθ +1) }{ (sinθ + cosθ -1)}$

   

$\tt   = \frac{ \bigg ( \dfrac {sinθ}{cosθ} + \dfrac{ cosθ}{cosθ} + \dfrac{1}{cosθ} \bigg) }{ \bigg( \dfrac{sinθ}{cosθ} + \dfrac{ cosθ}{cosθ }- \dfrac{1}{cosθ} \bigg)}$

     

$\tt   = \dfrac{(tanθ - 1 + secθ) }{( tanθ + 1 - secθ) }$

$\tt \: Since,$

$\tt \red \longmapsto\dfrac{ sinθ}{cosθ} =  tanθ \: \: \: and  \: \dfrac{1}{cosθ }= secθ$

   

$\tt \red \longmapsto   \dfrac {{ \{( tanθ + secθ) -1 \} }}{ {{ \{(tanθ - secθ) +1 \} }}}$

      Now multiplying both numerator and denominator by (tanθ - secθ), we get

   

$\tt \red \longmapsto     \dfrac{ { \{( tanθ + secθ) -1} \}}{ { \{{(tanθ - secθ) +1}} \}}$

$\tt \red \longmapsto \dfrac{\{( tanθ + secθ) -1 \} (tanθ - secθ)}{ \{(tanθ - secθ) +1 \}  (tanθ - secθ)}$

$\\\\ \tt\red\longmapsto \dfrac{\{(tan2θ - sec2θ) - (tanθ - secθ) \} }{ (tanθ - secθ + 1)(tanθ - secθ)}$

   

$\tt\red\longmapsto     \dfrac{(-1 -  \cancel{tanθ + secθ})}{( \cancel{tanθ - secθ +1})(tanθ - secθ)}$

$\tt\red\longmapsto     \dfrac{-1 }{ (tanθ - secθ)}$

$\tt\red\longmapsto    \dfrac{1}{ (secθ - tanθ)}$

       Hence, proved LHS = RHS

Answer 2

$\large\underline{\underline\blue{\sf We \: have : - }}$

sinθ−cosθ+1 / sinθ+cosθ−1 = 1/secθ−tanθ

$\large\underline{\underline\blue{\sf \: to \: find : - }}$

• prove that LHS = RHS

$\large \underline{ \underline{ \green{ \sf \: formula \: used : - }}}$

• sin²θ = 1 + tan²θ
• a² + b² = (a + b)(a - b)

$\large\underline{\underline\green{\sf \: solution : - }}$

we need to proved that,

$\tt \: \dfrac{sinθ - cosθ + 1}{sinθ + cosθ - 1} = \dfrac{1}{secθ - tanθ}$

we will use formula $\sf \red{ {sec}^{2}θ = 1 + {tan}^{2} θ }$

Now

$\sf \: \: \: \: LHS \: \: \: = \: \: \: \dfrac{sinθ- cosθ+1}{sinθ+cosθ - 1}$

Dividing numerator and denominator by cosθ , we get

$\tt \longrightarrow \: \dfrac{ \dfrac{sinθ - cosθ + 1}{cosθ}}{ \dfrac{sinθ + cosθ - 1}{cosθ} }$

Cancelling the similar terms in numerator and denominator we get,

$\tt \longrightarrow \: \: \dfrac{ \dfrac{sinθ - 1}{cosθ} + \dfrac{1}{cosθ} }{ \dfrac{sinθ + 1}{cosθ} - \dfrac{1}{cosθ} }$

we know that $\sf \red{tanθ = \dfrac{sinθ}{cosθ} } \:, \: \red{secθ = \dfrac{1}{cosθ} } \:$using these we get,

   

$\tt \longrightarrow \: \: \dfrac{tanθ - 1 + secθ}{tanθ + 1 - secθ}$

Multiplying both numerator and denominator by tanθ - secθ

$\tt \longrightarrow \: \: \dfrac{(tanθ + secθ + 1)(tanθ - secθ)}{(tanθ - secθ + 1)(tanθ - secθ)}$

Solving we get ,

$\tt \longrightarrow \: \: \dfrac{(tanθ + secθ)(tanθ - secθ) - 1(tanθ - secθ)}{(tanθ - secθ + 1)(tanθ - secθ)}$

Using the algebraic formula,

$\green{ \frak{ {a}^{2} - {b}^{2} = (a + b)(a - b) }}$

$\tt \longrightarrow \: \: \dfrac{ ({tan}^{2}θ - {sec}^{2} θ + 1) - (tanθ + sec θ)}{(tanθ - secθ)(tanθ - secθ)}$

Using the trigonometric formula, $\sf \red{sec ^{2} θ = 1 + {tan}^{2} θ}$

$\tt \longrightarrow \: \: \dfrac{ - 1 - tanθ + sec θ}{(tanθ - secθ + 1)(tan - secθ)}$

Taking- 1 common from numerator, we get

$\tt\: \longrightarrow \: \dfrac{ - \cancel{ (1tanθ - secθ)}}{ \cancel{(tanθ - secθ + 1) }{(tanθ - secθ)}}$

Cancelling the same term from numerator and denominator, we get

$\tt \longrightarrow \: \: \dfrac{1}{tanθ - secθ}$

Simplify we get,

$\tt \longrightarrow \: \: \dfrac{ 1 }{secθ - tanθ}$

$\tt = RHS$

therefore LHS = RHS

$\green{ \sf \: hence \: \: \: \dfrac{sinθ - cosθ + 1}{sinθ + cosθ - 1} = \dfrac{1}{secθ - tanθ} }$

$\qquad \qquad \qquad \qquad \huge \blue{ \underline \frak{hence \: proved}}$