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Answer:Step-by-step explanation:
QUESTION
\tt{ PROVE\ THAT \frac{cot A - cos A}{cot A + cos A}=\frac{cosec A - 1}{ cosec A + 1}}PROVE THATcotA+cosAcotA−cosA=cosecA+1cosecA−1
Taking L.H.S
\sf{= \frac{ cot A - cos A}{ cot A + cos A}}=cotA+cosAcotA−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}}→sinAcosA+cosAsinAcosA−cosA
\tt{→ \frac{ \frac{cos A - cos A sin A}{ sin A}}{\frac{cos A +cos A sin A}{ sin A}}}→sinAcosA+cosAsinAsinAcosA−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}}}→sinA1+sinAsinAsinA1−sinAsinA
\tt{→\frac{\frac{1}{sin A}-1}{\frac{1}{sin A}+1}}→sinA1+1sinA1−1
we\ know\ that ↓{\boxed{\sf{\red{\frac{1}{sin A}=cosec A }}}}we know that↓sinA1=cosecA
\sf{→\frac{cosec A –1}{cosec A + 1}= R.H.S.}→cosecA+1cosecA–1=R.H.S.
\tt{ \frac{cot A - cos A}{cot A + cos A}=\frac{cosec A - 1}{ cosec A + 1}}cotA+cosAcotA−cosA=cosecA+1cosecA−1
So,.
L.H.S = R.H.S
Hence pro
Answer:
Explanation:
sec
x
−
tan
x
=
1
cos
x
−
sin
x
cos
x
=
1
−
sin
x
cos
x
=
√
(
1
−
sin
x
)
2
√
cos
2
x
=
√
(
1
−
sin
x
)
2
√
1
−
sin
2
x
=
√
(
1
−
sin
x
)
2
1
−
sin
2
x
=
√
(
1
−
sin
x
)
2
(
1
+
sin
x
)
(
1
−
sin
x
)
=
√
1
−
sin
x
1
+
sin
x
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Step-by-step explanation:
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