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Answers
Answer:
UESTION
\tt{ PROVE\ THAT \frac{cot A - cos A}{cot A + cos A}=\frac{cosec A - 1}{ cosec A + 1}}PROVE THAT
cotA+cosA
cotA−cosA
=
cosecA+1
cosecA−1
Taking L.H.S
\sf{= \frac{ cot A - cos A}{ cot A + cos A}}=
cotA+cosA
cotA−cosA
\tt{\pink{Writing\ everything\ in\ terms\ of\ sin\ A\ and\ cos\ A}}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}}→
sinA
cosA
+cosA
sinA
cosA
−cosA
\tt{→ \frac{ \frac{cos A - cos A sin A}{ sin A}}{\frac{cos A +cos A sin A}{ sin A}}}→
sinA
cosA+cosAsinA
sinA
cosA−cosAsinA
\tt{→\frac{cos A (1- sin A)}{cos A (1 + sin A)}}→
cosA(1+sinA)
cosA(1−sinA)
\tt{→\frac{(1 - sin A)}{(1+ sin A)}}→
(1+sinA)
(1−sinA)
\sf{\red{Dividing\ sin\ A\ on\ numerator\ and\ denominator}}Dividing sin A on numerator and denominator
\tt{→ \frac{\frac{(1- sin A)}{ sin A}}{\frac{(1+sin A)}{ sin A}}}→
sinA
(1+sinA)
sinA
(1−sinA)
\tt{→\frac{\frac{1}{sin A}-\frac{sin A}{sin A}}{ \frac{1}{sin A}+\frac{sin A}{sin A}}}→
sinA
1
+
sinA
sinA
sinA
1
−
sinA
sinA
\tt{→\frac{\frac{1}{sin A}-1}{\frac{1}{sin A}+1}}→
sinA
1
+1
sinA
1
−1
we\ know\ that ↓{\boxed{\sf{\red{\frac{1}{sin A}=cosec A }}}}we know that↓
sinA
1
=cosecA
\sf{→\frac{cosec A –1}{cosec A + 1}= R.H.S.}→
cosecA+1
cosecA–1
=R.H.S.
\tt{ \frac{cot A - cos A}{cot A + cos A}=\frac{cosec A - 1}{ cosec A + 1}}
cotA+cosA
cotA−cosA
=
cosecA+1
cosecA−1
So,
L.H.S = R.H.S
Hence proved