Question

8. PS bisects ZQPR and PS I QR. If PQ = 2x
units, PR = (3y + 8) units, QS = x units and
SR = 2y units. Find the values of x and y.


​

Answer 1

Answer:

tex]\orange{\bold{\underbrace{\overbrace{❥Answer᎓}}}}[/tex]

Integrate the function

$\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}$

⇛$\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}$ ㅤ ㅤ ㅤ

⇛ $\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }$

⇛$\huge\tt {tan}^{ \frac{1}{2} - 1 } \times \frac{1}{ {cos}^{2} x}$ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x⇛(tan)$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)$

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$\bold\blue{☛\: Let tanx=t}$

$\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}$

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⇛$\huge\tt {sec}^{2} x = \frac{dt}{dx}$

⇛$\huge\tt{dx \frac{dt}{ {sec}^{2}x }}$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx$

⇛$\huge\tt ∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }$

⇛$\huge\tt ∫ {t}^{ - \frac{1}{2} }$ㅤ ㅤ

⇛ $\huge\tt\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 }$

⇛ $\huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}$

⇛$\huge2 \sqrt{t} + c = 2 \sqrt{tanx}$

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Answer 2

Answer:

Answer:

tex]\orange{\bold{\underbrace{\overbrace{❥Answer᎓}}}}[/tex]

Integrate the function

\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}

⇛\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}

sinxcosx

tanx

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}

sinxcosx×

cosx

cosx

tanx

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}

sinx×

cosx

cos

2

x

tanx

ㅤ ㅤ ㅤ

⇛ \huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }

cos

2

x×

cosx

sinx

tanx

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛\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

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛\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)

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛\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)

ㅤ ㅤ ㅤ ㅤ ㅤ

\bold\blue{☛\: Let tanx=t}☛Lettanx=t

\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}☛Differentiatingbothsidesw.r.t.x

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛\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

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛\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

ㅤ ㅤ

⇛ \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

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