Question

Maths brilliant. . .
Solve it fast. . .​

Answer 1

We have to prove that (sin A - cos A + 1)/(sin A + cos A - 1) = cos A/(1 - sin A)

Start from the left hand side

(sin A - cos A + 1)/(sin A + cos A - 1)

=> (sin A - cos A + 1)(sin A - cos A -1)/(sin A + cos A - 1)(sin A - cos A -1)

=> ((sin A - cos A)^2 - 1)/((sin A - 1)^2 - (cos A)^2)

=> ((sin A - cos A)^2 - 1)/((sin A)^2 - 2*sin A + 1 - (cos A)^2)

=> ((sin A - cos A)^2 - 1)/((sin A)^2 - 2*sin A + (sin A)^2)

=> ((sin A)^2 + (cos A)^2 - 2*sin A*cos A - 1)/((sin A)^2 - 2*sin A + (sin A)^2)

=> (1- 2*sin A*cos A - 1)/((sin A)^2 - 2*sin A + (sin A)^2)

=> (-2*sin A*cos A)/(2*(sin A)^2 - 2*sin A)

=> (-cos A)/(sin A - 1)

=> cos A/(1 - sin A)

which is the right hand side

This proves:(sin A - cos A + 1)/(sin A + cos A - 1) = cos A/(1 - sin A)

Answer 2

$\large\underline{\underline{\textbf{\textsf{\;To prove:-}}}}$

$\Longrightarrow\sf \dfrac{sin A-cos A+1}{sin A+cos A-1}=\dfrac{cos A}{1-sin A}$

$\large\underline{\underline{\textbf{\textsf{Solution:-}}}}$

$\Longrightarrow\sf \dfrac{sin A-cos A+1}{sin A+cos A-1}=\dfrac{cos A}{1-sin A}$

Multiply it by ( sin A - cos A - 1 ) in both numerator and denominator.

$\Longrightarrow\sf \dfrac{sin A-cos A+1}{sin A+cos A-1}\times \dfrac{sin A-cos A-1}{sin A-cos A-1}=\dfrac{cos A}{1-sin A}$

$\Longrightarrow\sf \dfrac{(sin A-cos A+1)( sin A-cos A-1)}{(sin A+cos A-1)(sin A-cos A-1)}=\dfrac{cos A}{1-sin A}$

$\Longrightarrow\sf \dfrac{[(sin A-cos A)+1][( sin A-cos A)-1]}{[(sin A-1)+cos A][(sin A-1)-cos A]}=\dfrac{cos A}{1-sin A}$

:: Applying identity:-

• (A+B)(A-B)=A²-B²

$\Longrightarrow\sf \dfrac{(sin A-cos A)^2-(1)^2}{(sin A-1)^2-cos^2 A}=\dfrac{cos A}{1-sin A}$

:: Applying identity:-

• (A-B)²=A²+B²-2AB

$\Longrightarrow\sf \dfrac{sin^2 A+cos^2 A-2 (sin A )(cos A)-1}{sin^2 A+1-2(sin A)(1)-cos^2A}=\dfrac{cos A}{1-sin A}$

$\Longrightarrow\sf \dfrac{sin^2 A+cos^2 A-2 sin A cos A-1}{sin^2 A-2sin A+(1-cos^2A)}=\dfrac{cos A}{1-sin A}$

:: Applying identity:-

• sin²A+cos²A=1

• 1-cos²A=sin²A

$\Longrightarrow\sf \dfrac{1-2 sin A cos A-1}{sin^2 A-2sin A+(sin^2A)}=\dfrac{cos A}{1-sin A}$

$\Longrightarrow\sf \dfrac{-2 sin A cos A}{2sin^2 A-2sin A}=\dfrac{cos A}{1-sin A}$

$\Longrightarrow\sf \dfrac{-2 sin A cos A}{2sinA(sinA -1)}=\dfrac{cos A}{1-sin A}$

:: Cancel 2sin A from both numerator and denominator

$\Longrightarrow\sf \dfrac{- cos A}{(sinA -1)}=\dfrac{cos A}{1-sin A}$

:: Taking -1 as common

$\Longrightarrow\sf \dfrac{-1( cos A)}{-1(1-sinA )}=\dfrac{cos A}{1-sin A}$

:: Cancel -1

$\Longrightarrow\sf \dfrac{cos A}{1-sinA }=\dfrac{cos A}{1-sin A}$

Hence proved LHS=RHS