Prove that if the sum of the sequences of the roots of the equation ax²+bx+c=0is 1 then b²=2ac+a²
Answers
Step-by-step explanation:
Integrate the function
\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}
⇛\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}
sinxcosx
tanx
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⇛\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}
sinxcosx×
cosx
cosx
tanx
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⇛\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}
sinx×
cosx
cos
2
x
tanx
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⇛ \huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }
cos
2
x×
cosx
sinx
tanx
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⇛\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }
cos
2
x×tanx
tanx
⇛\huge\tt {tan}^{ \frac{1}{2} - 1 } \times \frac{1}{ {cos}^{2} x}tan
2
1
−1
×
cos
2
x
1
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⇛\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x⇛(tan)(tan)
−
2
1
×
cos
2
x
1
=(tanx)
−
2
1
×sec
2
x⇛(tan)
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⇛\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)(tan)
−
2
1
×
cos
2
x
1
=∫(tanx)
−
2
1
×sec
2
x×dx⇛(tan)
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\bold\blue{☛\: Let tanx=t}☛Lettanx=t
\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}☛Differentiatingbothsidesw.r.t.x
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⇛\huge\tt {sec}^{2} x = \frac{dt}{dx}sec
2
x=
dx
dt
⇛\huge\tt{dx \frac{dt}{ {sec}^{2}x }}dx
sec
2
x
dt
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⇛\huge\tt∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx∴∫(tanx)
−
2
1
×sec
2
x×dx
⇛\huge\tt ∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }∫(t)
−
2
1
×sec
2
x×
sec
2
x
dt
⇛\huge\tt ∫ {t}^{ - \frac{1}{2} }∫t
−
2
1
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⇛ \huge\tt\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 }
−
2
1
+1
t
−
2
1
+1
⇛ \huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}
2
1
t
2
1
+c=2t
2
1
+c=2
t
⇛\huge2 \sqrt{t} + c = 2 \sqrt{tanx}2
t
+c=2
tanx
It may help you okk my dear friend
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