Math, asked by nalini81g, 6 months ago

sin 0 ÷
1+cos 0
+
1+cos 0÷ sin 0
2 cos ec0​

Answers

Answered by amirmerani786
0

Answer:

Step-by-step explanation:

\green{\bold{\underline{ ✪ UPSC-ASPIRANT✪ }}}

✪UPSC−ASPIRANT✪

\red{\bold{\underline{\underline{QUESTION:-}}}}

QUESTION:−

Q:-solve and verify the equation

\frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{ 3} )

3

1

x−4=x−(

2

1

+

3

x

)

\huge\tt\underline\blue{Answer }

Answer

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

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⟹ \frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{3} )⟹

3

1

x−4=x−(

2

1

+

3

x

)

⟹ \frac{x}{3} - 4 = x - ( \frac{3 + 2x}{6} )⟹

3

x

−4=x−(

6

3+2x

)

⟹ \frac{x - 12}{3} = x - ( \frac{2x + 3}{6} )⟹

3

x−12

=x−(

6

2x+3

)

⟹ \frac{x - 12}{3} = x - \frac{2x - 3}{6}⟹

3

x−12

=x−

6

2x−3

⟹ \frac{x - 12}{3} = \frac{6x - 2x - 3}{6}⟹

3

x−12

=

6

6x−2x−3

⟹ \frac{x - 12}{3} = \frac{4x - 3}{6}⟹

3

x−12

=

6

4x−3

cancelling 6( R.H.S) By 3 From L.H.S

⟹ \frac{x - 12}{1} = \frac{4x - 3}{2}⟹

1

x−12

=

2

4x−3

⟹ 2(x - 12) = 4x - 3⟹2(x−12)=4x−3

⟹ 2x - 24 = 4x - 3⟹2x−24=4x−3

⟹ - 24 + 3 = 4x - 2x⟹−24+3=4x−2x

⟹ - 21 = 2x⟹−21=2x

⟹ x = - \frac{21}{2}⟹x=−

2

21

CHECK:-

⟹ \frac{ - \frac{21}{2} }{3} - 4 = - \frac{21}{2} - ( \frac{1}{2} + ( - ) \frac{ \frac{21}{2} }{3} )⟹

3

2

21

−4=−

2

21

−(

2

1

+(−)

3

2

21

)

⟹ - \frac{21}{6} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{21}{6} )⟹−

6

21

−4=−

2

21

−(

2

1

6

21

)

⟹ - \frac{7}{2} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{7}{2} )⟹−

2

7

−4=−

2

21

−(

2

1

2

7

)

⟹ \frac{ - 7 - 8}{2} = - \frac{21}{2} - ( - \frac{6}{2} )⟹

2

−7−8

=−

2

21

−(−

2

6

)

⟹ - \frac{15}{2} = - \frac{21}{2} - ( - 3)⟹−

2

15

=−

2

21

−(−3)

⟹ - \frac{15}{2} = - \frac{21}{2} + 3⟹−

2

15

=−

2

21

+3

⟹ - \frac{15}{2} = \frac{ - 21 + 6}{2} = - \frac{15}{2}⟹−

2

15

=

2

−21+6

=−

2

15

THEREFORE,L.H.S=R.H.S

VERIFIED✔️

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HOPE IT HELPS YOU..

_____________________

Thankyou:)

Answered by rajeevr06
1

Answer:

 \frac{sin \alpha }{1 + cos \alpha }  +  \frac{1 + cos \alpha }{sin \alpha }  =  \frac{ {sin}^{2}  \alpha  +  {cos}^{2} \alpha  + 1 + 2cos \alpha  }{sin \alpha (1 + cos \alpha )}

 \frac{2 + 2cos \alpha }{sin \alpha (1 + cos \alpha )}  =  \frac{2}{sin \alpha }  = 2cosec \alpha

RHS. Proved

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