Question

Prove That Trinometry. Question is below :

Answer 1

To prove:

$\bf{\dfrac{sin A - cos A + 1}{sin A + cosA - 1}= \dfrac{1 + sin A}{cos A}}$

Solution:

$\sf{Dividing \: by \: cosA \: on \: both \: sides,}\\\\\\\sf{Taking \: LHS,}\\\\\\\sf{\dfrac{\dfrac{sinA}{cosA} - \dfrac{cos A}{cosA}+ \dfrac{1}{cos A}}{\dfrac{sinA}{cosA}+ \dfrac{cosA}{cosA}-\dfrac{1}{cos A}}}$

$\\\\= \sf{\dfrac{tanA-1+secA}{tanA+1-secA}}\\\\\\\\= \sf{\dfrac{secA+tanA-1}{tanA-secA+1}}$

$\\\\\rm{1 \: can \: be \: written \: as \: (sec^{2} A - tan^{2} A)}\\\\\\= \sf{\dfrac{secA+tanA-(sec^{2} A - tan^{2} A)}{tan A - secA + 1}}$

Split sec²A - tan²A Using a² - b²= (a + b)(a - b)

$\\\\= \sf{\dfrac{(sec A+tanA) - [(secA - tanA)(secA+tanA)]}{tan A - secA + 1}}$

Taking out secA + tanA as a common factor,

$\\\\= \sf{\dfrac{(secA+tanA)(1-secA+tanA)}{1 - secA + tanA}}\\\\\\\\= \sf{secA + tan A}\\\\\\= \sf{\dfrac{1}{cosA}+\dfrac{sinA}{cosA}}\\\\\\\\= \sf{\dfrac{1 + sinA}{cosA}}\\\\\\\boxed{\purple{\bf{Hence \: proved}}}$