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$$ \frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A) $$

\begin{align} \mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px] &= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))} \\[6px] &= \frac{2+2\sin(A)\cos(A)}{\sin(A)(1+\cos(A))} \end{align}

What should be done from here?

egreg
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user366082
  • 437

7 Answers7

2

we have $$\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=\frac{\sin^2(A)+(1+\cos(A))^2}{\sin(A)(1+\cos(A))}=\frac{2+2\cos(A)}{\sin(A)(1+\cos(A))}=$$ can you finish?

Dr. Sonnhard Graubner
  • 95,283
2

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

Now, usse $$\sin^2(A) + \cos^2(A) = 1.$$

Take it from here.

amWhy
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Dhruv Kohli
  • 5,216
  • Very nicely done! Just tweaked the formatting with the backslash immediately before any trig-function. I sincerely think this is a very helpful answer! – amWhy Jun 28 '17 at 15:06
2

enter image description here[take 2 common in the numerator then you will get 2(1+cos(A))]

[in denominator there is a mistake it's 1+cos(A)]

ashok knv
  • 103
2

You make several mistakes, the main one being $$ (a+b)^2=a^2+b^2 $$

The mistake is $(1+\cos(A))^2=1+\cos^2(A)$, whereas it should be $$ (1+\cos(A))^2=1+2\cos(A)+\cos^2(A) $$

Note that $$ \frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab} $$ where $a=\sin(A)$ and $b=1+\cos(A)$.

In the second step you also arbitrarily insert a term $2\sin(A)\cos(A)$ with no justification.

Start again.

egreg
  • 238,574
2

$$\frac{\sin{A}}{1+\cos{A}}+\frac{1+\cos{A}}{\sin{A}}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2}}+\frac{2\cos^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}=$$ $$=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}+\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}=\frac{\sin^2\frac{A}{2}+\cos^2\frac{A}{2}}{\sin\frac{A}{2}\cos\frac{A}{2}}=\frac{2}{2\sin\frac{A}{2}\cos\frac{A}{2}}=\frac{2}{\sin{A}}.$$

Michael Rozenberg
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}} \newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}$

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

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

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

$$\frac{2+2\cos a}{\sin a (1+ \cos a)} = \frac{2}{\sin a}$$

$$\bbx{\frac{1+1\cos a}{1 (1+ \cos a)} = \frac{1}{1}}$$

MCCCS
  • 1,625
1

Alternatively: $$\frac{\sin A(1-\cos A)}{\sin^2 A}+\frac{1+\cos A}{\sin A}=2cosec A.$$

farruhota
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