Question

sin - cos + 1 /sin + cos + 1 = 1 /sec- tan​

Answer 1

${\boxed{\underline{\tt{ \orange{Required \: \: Answer:-}}}}}$

◒TO PROVE:-

$\tt {\frac{sin \theta \: - \: cos \theta + 1}{ sin \theta \: + \: cos \theta - 1\: } = \frac{1}{sec \theta \: - \: tan \theta} }$

◒USED:-

• $\bf{ {sec}^{2} \theta \: = 1 + {tan}^{2} \theta \: }$
• $\bf{ {sec}^{2} \theta \: - {tan}^{2} \theta = 1}$
• ${tan}^{2} \theta \: - {sec}^{2} \theta = - 1$
• $(a - b)(a + b) = {a}^{2} - {b}^{2}$

◒SOLUTION:-

${\boxed{\underline{\mathbb{\purple{ L.H.S}}}}}$

$: \longrightarrow \tt {\frac{sin \theta \: - \: cos \theta + 1}{ sin \theta \: + \: cos \theta - 1\: } }$

Dividing numerator and denominator by cosθ

$: \longrightarrow \tt{ \frac{ \frac{1}{cos \theta} (sin \theta - cos \theta + 1)}{ \frac{1}{cosθ}(sin \theta + cos \theta - 1) } }$

$: \longrightarrow \tt \frac{{( \frac{sin \theta}{cos \theta} ) - (\frac{cos \theta}{cos \theta}) + ( \frac{1}{cos \theta} ) }}{( \frac{sin \theta}{cos \theta}) + (\frac{cos \theta}{cos \theta} ) - ( \frac{1}{cos \theta} )}$

$: \longrightarrow \tt{ \frac{tan \theta - 1 + sec \theta}{tan \theta + 1 - sec \theta}}$

$: \longrightarrow \tt{ \frac{tan \theta + sec \theta - 1}{tan \theta - sec \theta + 1} }$

Multiplying both Numerator and Denominator by

(tanθ-secθ).

$: \longrightarrow \tt{ \frac{{(tan \theta + sec \theta - 1)(tan \theta - sec \theta)}}{ \: {(tan \theta - sec \theta + 1})(tan \theta - sec \theta) } }$

$: \longrightarrow \tt{ \frac{(tan \theta + sec \theta)(tan \theta - sec \theta) - 1 \times (tan \theta - sec \theta)}{(tan \theta - sec \theta + 1)(tan \theta - sec \theta)}}$

Using (a-b) (a+b)= a^2 - b^2

Here,

a = tanθ and b = secθ

$: \longrightarrow \tt{ \frac{( {tan}^{2} \theta - {sec}^{2} \theta) - (tan \theta - sec \theta) }{(tan \theta - sec \theta + 1)(tan \theta - sec \theta)} }$

$: \longrightarrow \tt{ \frac{( - 1) - (tan \theta - sec \theta)}{(tan \theta - sec \theta + 1)(tan \theta - sec \theta)} }$

$: \longrightarrow \tt{ \frac{ - 1 - tan \theta + sec \theta}{(tan \theta - sec \theta + 1)(tan \theta - sec \theta)} }$

$: \longrightarrow \tt{ \frac{- (1 + tan \theta + sec \theta)}{(tan \theta - sec \theta + 1)(tan \theta - sec \theta)} }$

$: \longrightarrow \tt{ \frac{ - 1}{tan \theta - sec \theta} }$

$: \longrightarrow \tt{ \frac{1}{ - (tan \theta - sec \theta)} }$

$: \longrightarrow \tt{ \frac{1}{sec \theta - tan \theta} }$

${\boxed{\underline{\mathbb{\purple{R .H.S}}}}}$

Hence proved.