Question

prove that (cot A - cos A)÷(cot A +cos A)=(cosecA-1)÷(codec A+1)​

Answer 1

simplify in terms of cos and sin and taking common you will get simple form and divide numerator a d denominator by sin a and get the ans

Answer 2

$\tt{\red{QUESTION}}$

$\tt{ PROVE\ THAT \frac{cot A - cos A}{cot A + cos A}=\frac{cosec A - 1}{ cosec A + 1}}$

Taking L.H.S

$\sf{= \frac{ cot A - cos A}{ cot A + cos A}}$

$\tt{\pink{Writing\ everything\ in\ terms\ of\ sin\ A\ and\ cos\ A}}$

$\tt{→ \frac{ \frac{cos A}{ sin A} -cos A}{\frac{cos A}{sin A} +cos A}}$

$\tt{→ \frac{ \frac{cos A - cos A sin A}{ sin A}}{\frac{cos A +cos A sin A}{ sin A}}}$

$\tt{→\frac{cos A (1- sin A)}{cos A (1 + sin A)}}$

$\tt{→\frac{(1 - sin A)}{(1+ sin A)}}$

$\sf{\red{Dividing\ sin\ A\ on\ numerator\ and\ denominator}}$

$\tt{→ \frac{\frac{(1- sin A)}{ sin A}}{\frac{(1+sin A)}{ sin A}}}$

$\tt{→\frac{\frac{1}{sin A}-\frac{sin A}{sin A}}{ \frac{1}{sin A}+\frac{sin A}{sin A}}}$

$\tt{→\frac{\frac{1}{sin A}-1}{\frac{1}{sin A}+1}}$

$we\ know\ that ↓{\boxed{\sf{\red{\frac{1}{sin A}=cosec A }}}}$

$\sf{→\frac{cosec A –1}{cosec A + 1}= R.H.S.}$

$\tt{ \frac{cot A - cos A}{cot A + cos A}=\frac{cosec A - 1}{ cosec A + 1}}$

So,

L.H.S = R.H.S

Hence proved