Question

Prove: $\frac{1+secA}{secA}=\frac{si{n}^{2}A}{1-cosA}$

Answer 1

$L.H.S.=\frac{1+secA}{secA}$

$=\frac{1+\frac{1}{cosA}}{\frac{1}{cosA}}$ ............... $\sec x=\frac{1}{\cos x}$

$=\frac{\frac{1+cosA}{cosA}}{\frac{1}{cosA}}$

$=\frac{1+cosA}{1}$

Now multiply the numerator and denominator with $1-cosA$

$=\frac{(1+cosA)(1-cosA)}{1-cosA}$

$=\frac{1-co{s}^{2}A}{1-cosA}$ ........... $a^2-b^2=(a-b)(a+b)$

$=\frac{si{n}^{2}A}{1-cosA}$ .............. $[sin²\theta +cos²\theta =1]$

$=R.H.S.$