Question

Prove that sinA-cosA+1÷sinA+cosA-1 = 1÷secA-tanA

Answer 1

Answer:

your answer attached in the photo

Answer 2

$\displaystyle \sf{ \frac{ \sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{1}{ \sec A \: - \tan A} } \:is \: \:proved$

Given :

$\displaystyle \sf{ \frac{ \sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{1}{ \sec A \: - \tan A} }$

To find :

To prove the expression

Solution :

Step 1 of 2 :

Write down the given expression

Here the given expression is

$\displaystyle \sf{ \frac{ \sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{1}{ \sec A \: - \tan A} }$

Step 2 of 2 :

Prove the expression

LHS

$\displaystyle \sf{ = \frac{ \sin A - \cos A + 1}{\sin A + \cos A - 1} }$

Dividing numerator and denominator both by cos A we get

$\displaystyle \sf{ = \frac{ \tan A - 1 + \sec A }{\tan A + 1 - \sec A } }$

$\displaystyle \sf{ = \frac{ \tan A - 1 + \sec A }{\tan A - \sec A + 1 } }$

$\displaystyle \sf{ = \frac{ \tan A - 1 + \sec A }{(\tan A - \sec A ) - ( { \tan}^{2} A - { \sec}^{2} A ) } }$

$\displaystyle \sf{ = \frac{ \tan A - 1 + \sec A }{(\tan A - \sec A ) - (\tan A + \sec A ) (\tan A - \sec A ) } }$

$\displaystyle \sf{ = \frac{ \tan A - 1 + \sec A }{(\tan A - \sec A ) (1 - \tan A - \sec A ) } }$

$\displaystyle \sf{ = \frac{ \tan A + \sec A - 1 }{(\sec A - \tan A ) (\tan A + \sec A - 1 ) } }$

$\displaystyle \sf{ = \frac{1 }{\sec A - \tan A } }$

= RHS

Hence proved

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