Question

sin/1+cosA+1+cosA/sinA

Answer 1

Hey !!!

sinA / 1 + cosA + 1 + cosA/ sinA

=> sin² A + (1 + cosA)²
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sinA ( 1 + cosA)

=> sin²A + 1 + cos²A + 2cosA
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sinA ( 1 + cosA )

=> sin²A+ cos²A + 1 + 2cosA
------------------------------------------------
sinA ( 1 + cos A)

=> 2 + 2cosA
-----------------------------
sinA ( 1 + cosA )

=> 2( 1 + cosA)
----------------------
sinA ( 1+ cosA)

=> 2/sinA = 2cosecA RHS prooved

Hope it helps you !!!

@Rajukumar111###

Answer 2

Solution:

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Given & To find:


$\frac{sinA}{cosA+1}+ \frac{cosA+1}{sin A}$ = ?


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As we know,

[tex] \frac{cosA+1}{sinA} + \frac{sinA}{cosA+1} [/tex]

= $\frac{(cosA+1)(cosA+1)+(sinA)(sinA)}{(sinA)(cosA+1)}$    (by cross-multiplication)

= $= \frac{(cosA+1)^2+(sinA)^2}{(sinA)(cosA+1)}$

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As we know the formula that,


(a + b)² = a² + 2ab + b²


= $\frac{cos^2A+2cosA+1+sin^2A}{(sinA)(cosA+1)}$


= $\frac{sin^2A+cos^2A + 2cosA +1 }{sinA(1+cosA)}$


we know that,


$sin^2A + cos^2A = 1$


so,


=$\frac{1+2cosA+1}{sinA(1+cosA)}$


=$\frac{2+2cosA}{sinA(1+cosA)}$


=$\frac{2(1+cosA)}{sinA(1+cosA)}$

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By, canceling cosA + 1 , as it is in both numerator and denominator
we get,


= $\frac{2}{sinA}$


=$2 ( \frac{1}{sinA})$


We know that,


$\frac{1}{sinA} = cosecA$


So, substituting $\frac{1}{sinA} = cosecA$,

We get,

= $2cosecA$ ,.. = RHS

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                                                        Hope it Helps..