How to solve this identity? Thank you!

Question

How to solve this identity? Thank you!

$\frac{1}{sin a} + \frac{1}{tan a} = cot (\frac{a}{2})$ $1/sina + 1/tana = cot(a/2)$

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Answer 1 · 2017-12-10T12:42:52

If we know our trigonometric inverses right, we can simplify as follows:
$\frac{1}{sin (a)} = csc (a)$ $1/sin(a)=csc(a)$
$\frac{1}{tan (a)} = cot (a)$ $1/tan(a)=cot(a)$

$\frac{1}{sin (a)} + \frac{1}{tan (a)} = csc (a) + cot (a)$ $1/sin(a)+1/tan(a)=csc(a)+cot(a)$

This is just the result of the half angle formula for $cot$ $cot$ :
$cot (\frac{θ}{2}) = csc (θ) + cot (θ)$ $cot(theta/2)=csc(theta)+cot(theta)$

So we can simplify to:
$= cot (\frac{a}{2})$ $=cot(a/2)$ , which is what we wanted to prove.

Answer 2 · 2017-12-10T12:53:50

See explanation.

Explanation:

$L = \frac{1}{sin α} + \frac{1}{tan α} = \frac{1}{sin α} + \frac{cos α}{sin α} =$ $L=1/sinalpha+1/tanalpha=1/sinalpha+cosalpha/sinalpha=$

$= \frac{1 + cos α}{sin α} = \frac{1 + cos 2 \cdot (\frac{α}{2})}{sin (2 \cdot \frac{α}{2})} =$ $=(1+cosalpha)/sinalpha=(1+cos2*(alpha/2))/sin(2*alpha/2)=$

$\frac{1 + 2 {cos}^{2} (\frac{α}{2}) - 1}{2 sin (\frac{α}{2}) cos (\frac{α}{2})} =$ $(1+2cos^2(alpha/2)-1)/(2sin(alpha/2)cos(alpha/2))=$

$\frac{2 {cos}^{2} (\frac{α}{2})}{2 sin (\frac{α}{2}) cos (\frac{α}{2})} =$ $(2cos^2(alpha/2))/(2sin(alpha/2)cos(alpha/2))=$

$(cos^(cancel(2)}(alpha/2))/(sin(alpha/2)cancel(cos(alpha/2)))=$

$(cos(alpha/2))/(sin(alpha/2))=cot(alpha/2)=R$

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How to solve this identity? Thank you!

$\frac{1}{sin a} + \frac{1}{tan a} = cot (\frac{a}{2})$ $1/sina + 1/tana = cot(a/2)$

2 Answers

Explanation:

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Science

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How to solve this identity? Thank you!

1/sina + 1/tana = cot(a/2)1sina+1tana=cot(a2)1/sina + 1/tana = cot(a/2)

2 Answers

Explanation:

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$\frac{1}{sin a} + \frac{1}{tan a} = cot (\frac{a}{2})$ $1/sina + 1/tana = cot(a/2)$