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Prove That:-
Answer:
Prove That:-
\large \sf \cot^{ - 1} ( \frac{ \sqrt{1 + \sin \: x } + \sqrt{1 - \sin x} }{ \sqrt{1 - \sin \: x } - \sqrt{1 - \sin \: x } } = \frac{x}{2} ;∈(0, \frac{\pi}{4} )cot
−1
(
1−sinx
−
1−sinx
1+sinx
+
1−sinx
=
2
x
;∈(0,
4
π
)
\huge \sf{ \underline{ \color{maroon}{Solution:- }}}
Solution:−
\large \sf \: L.H.S = \ \cot^{ - 1} ( \frac{ \sqrt{1 + \sin \: x } + \sqrt{1 - \sin x} }{ \sqrt{1 - \sin \: x } - \sqrt{1 - \sin \: x } }L.H.S= cot
−1
(
1−sinx
−
1−sinx
1+sinx
+
1−sinx
\sf \: ⇒ \cot ^{ - 1} ( \frac{ \sqrt{1 + \sin \:x } \: + \sqrt{1 - \sin \: x} }{ \sqrt{1 + \sin \: x } - \sqrt{1 - \sin \: x } } \times \frac{( \sqrt{1 + \sin \: x } + \sqrt{1 + \sin \: x }) }{( \sqrt{1 + \sin \: x } + \sqrt{1 - \sin \: x } }⇒cot
−1
(
1+sinx
−
1−sinx
1+sinx
+
1−sinx
×
(
1+sinx
+
1−sinx
(
1+sinx
+
1+sinx
)
⇒ \cot ^{ - 1} ( \frac{1 + \sin \: x + 1 - \sin \: x + 2 \sqrt{1 - \sin \: x } }{1 + \sin \: x - 1 + \sin \: x } )⇒cot
−1
(
1+sinx−1+sinx
1+sinx+1−sinx+2
1−sinx
)
⇒ \cot ^{ - 1} ( \frac{2 + 2 \cos \: x}{2 \sin \: x } )⇒cot
−1
(
2sinx
2+2cosx
)
\boxed{ \color{maroon}{⇒ \cot^{ - 1} ( \frac{1 + \cos \: x }{ \sin \: x } )}}
⇒cot
−1
(
sinx
1+cosx
)
⇒ \bf\cot ^{ - 1} ( \frac{2 \cos \: x \frac{x}{2} }{2 \sin \frac{x}{2} \cos \frac{x}{2} } )⇒cot
−1
(
2sin
2
x
cos
2
x
2cosx
2
x
)
⇒ \cot ^{ - 1}( { \cot}{ \frac{x}{2} } )⇒cot
−1
(cot
2
x
)
\boxed{ \color{goldenrod}{⇒ \frac{x}{2} = \bf \: R.H.S}}
⇒
2
x
=R.H.S
\huge \pink \dag \: { \large{ \boxed{ \red{Hence \: proved}}}}†
Henceproved
Step-by-step explanation:
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