Owning up and being responsible for what you do and are able to give a satisfactory reason for it, is known as- *
Accountability
Dependability
Both the above
None of the above
Answers
Answer:
_______________don't know
Explanation:
TO PROVE:
\sf cot ^{ - 1} \left(\dfrac{ \sqrt{1 + \sin \: x } + \sqrt{1 - \sin \: x } }{ \sqrt{1 + \sin \: x} - \sqrt{1 - \sin \: x } } \right )= \dfrac{x}{2}cot
−1
(
1+sinx
−
1−sinx
1+sinx
+
1−sinx
)=
2
x
PROOF:
\leadsto\sf Take \ \sqrt{1 + sin\ x}⇝Take
1+sin x
\boxed{ \large{ \bold{ \gray{sin \ 2x = 2 \ sin \ x \ cos \ x}}}}
sin 2x=2 sin x cos x
\leadsto \sf sin \ x = 2 \ sin \ \left(\dfrac{x}{2}\right) \ cos \ \left( \dfrac{x}{2} \right)⇝sin x=2 sin (
2
x
) cos (
2
x
)
Add one on both sides.
\leadsto\sf 1 + sin \ x = 1 + 2 \ sin \ \left(\dfrac{x}{2}\right) \ cos \ \left( \dfrac{x}{2} \right)⇝1+sin x=1+2 sin (
2
x
) cos (
2
x
)
\boxed{ \large{ \bold{ \gray{sin ^{2} \ \frac{x}{2} + {cos}^{2} \ \frac{x}{2} =1}}}}
sin
2
2
x
+cos
2
2
x
=1
1 + 2 sin(x/2) cos (x/2) = sin² (x/2) + cos² (x/2) + 2 sin(x/2) cos (x/2)
\boxed{ \large{ \bold{ \gray{ A^2+ 2AB + B^2 = (A+B)^2}}}}
A
2
+2AB+B
2
=(A+B)
2
1 - sin x = [sin (x/2) - cos (x/2)]²
\leadsto\sf 1 + sin \ x = \left( sin \ \left(\dfrac{x}{2}\right) + \ cos \ \left( \dfrac{x}{2} \right) \right) ^{2}⇝1+sin x=(sin (
2
x
)+ cos (
2
x
))
2
\leadsto\sf \sqrt{1 + sin \ x} = sin \ \left(\dfrac{x}{2}\right) + \ cos \ \left( \dfrac{x}{2} \right)⇝
1+sin x
=sin (
2
x
)+ cos (
2
x
)
\sf Take \ \sqrt{1 - sin\ x}Take
1−sin x
\boxed{ \large{ \bold{ \gray{sin \ 2x = 2 \ sin \ x \ cos \ x}}}}
sin 2x=2 sin x cos x
\leadsto \sf - sin \ x = - 2 \ sin \ \left(\dfrac{x}{2}\right) \ cos \ \left( \dfrac{x}{2} \right)⇝−sin x=−2 sin (
2
x
) cos (
2
x
)
Add one on both sides
\leadsto\sf 1 - sin \ x = 1 - 2 \ sin \ \left(\dfrac{x}{2}\right) \ cos \ \left( \dfrac{x}{2} \right)⇝1−sin x=1−2 sin (
2
x
) cos (
2
x
)
\boxed{ \large{ \bold{ \gray{sin ^{2} \ \frac{x}{2} + {cos}^{2} \ \frac{x}{2} =1}}}}
sin
2
2
x
+cos
2
2
x
=1
1 - 2 sin(x/2) cos (x/2) = sin² (x/2) + cos² (x/2) - 2 sin(x/2) cos (x/2)
\boxed{ \large{ \bold{ \gray{ A^2 - 2AB + B^2 = (A - B)^2}}}}
A
2
−2AB+B
2
=(A−B)
2
1 - sin x = [sin (x/2) - cos (x/2)]²
This can also be written as:
1 - sin x = [cos (x/2) - sin (x/2)]²
\leadsto\sf \sqrt{1 - sin \ x }= \ cos \ \left( \dfrac{x}{2} \right) - sin \ \left(\dfrac{x}{2}\right)⇝
1−sin x
= cos (
2
x
)−sin (
2
x
)
\leadsto \sf Take \ \sqrt{1 + sin\ x} + \sqrt{1 - sin\ x}⇝Take
1+sin x
+
1−sin x
sin (x/2) + cos (x/2) + cos (x/2) - sin (x/2)
\leadsto\sf \sqrt{1 + sin\ x} + \sqrt{1 - sin\ x} = 2 cos \ \left(\dfrac{x}{2} \right)⇝
1+sin x
+
1−sin x
=2cos (
2
x
)
\sf Take \ \sqrt{1 + sin\ x} - \sqrt{1 - sin\ x}Take
1+sin x
−
1−sin x
sin (x/2) + cos (x/2) - cos (x/2) + sin (x/2)
\leadsto\sf \sqrt{1 + sin\ x} - \sqrt{1 - sin\ x} = 2 sin \ \left(\dfrac{x}{2} \right)⇝
1+sin x
−
1−sin x
=2sin (
2
x
)
Substitute the respective values.
\leadsto\sf \cot ^{ - 1} \left( \dfrac{2cos \ \left(\dfrac{x}{2} \right )}{ 2sin \ \left(\dfrac{x}{2}\right ) }\right )⇝cot
−1
⎝
⎜
⎛
2sin (
2
x
)
2cos (
2
x
)
⎠
⎟
⎞
\leadsto\sf \cancel{cot}^{ - 1} \left( \cancel{cot} \ \left( \dfrac{x}{2} \right)\right)⇝
cot
−1
(
cot
(
2
x