Using $t = tan (\frac{1}{2} θ)$ $t=tan(1/2theta)$ , how do you prove that $\frac{tan 2 α + cot α}{tan 2 α - tan α} = {cot}^{2} α$ $(tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alpha$ ?

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Using $t = tan (\frac{1}{2} θ)$ $t=tan(1/2theta)$ , how do you prove that $\frac{tan 2 α + cot α}{tan 2 α - tan α} = {cot}^{2} α$ $(tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alpha$ ?

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Answer 1 · 2018-05-30T11:35:45

We use $tan (2 x) = 2 \frac{tan (x)}{1 - {tan}^{2} (x)}$ $tan(2x)=2tan(x)/(1-tan^2(x))$

Explanation:

Using the formula above we get
$\frac{2 \frac{tan (x)}{1 - {tan}^{2} (x)} + \frac{1}{tan (x)}}{2 \frac{tan (x)}{1 - {tan}^{2} (x)} - - tan (x)}$ $(2tan(x)/(1-tan^2(x))+1/tan(x))/(2tan(x)/(1-tan^2(x))--tan(x))$
Multiplying numerator and denominator by
$1 - {tan}^{2} (x)$ $1-tan^2(x)$

we get

$\frac{2 tan (x) + (\frac{1 - {tan}^{2} (x)}{tan (x)})}{2 tan (x) - tan (x) (1 - {tan}^{2} (x))}$ $(2tan(x)+((1-tan^2(x))/tan(x)))/(2tan(x)-tan(x)(1-tan^2(x))$
multiplying numerator and denominator by

$tan (x)$ $tan(x)$
$\frac{2 {tan}^{2} (x) + 1 - {tan}^{2} (x)}{2 {tan}^{2} (x) - {tan}^{2} (x) (1 - {tan}^{2} (x))}$ $(2tan^2(x)+1-tan^2(x))/(2tan^2(x)-tan^2(x)(1-tan^2(x))$
and this is

$\frac{{tan}^{2} (x) + 1}{{tan}^{2} (x) (1 + {tan}^{2} (x))} = {cot}^{2} (x)$ $(tan^2(x)+1)/(tan^2(x)(1+tan^2(x)))=cot^2(x)$

Answer 2 · 2018-05-30T12:02:36

Please see below.

Explanation:

We know that,

$(1) cos (A - B) = cos A cos B + sin A sin B$ $color(red)((1)cos(A-B)=cosAcosB+sinAsinB$

$(2) sin (A - B) = sin A cos B - cos A sin B$ $color(blue)((2)sin(A-B)=sinAcosB-cosAsinB$

Here,

$\frac{tan 2 α + cot α}{tan 2 α - tan α} = {cot}^{2} α$ $(tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alpha$

We take,

$L H S = \frac{tan 2 α + cot α}{tan 2 α - tan α}$ $LHS=(tan2alpha+cotalpha)/(tan2alpha-tanalpha)$

$L H S = \frac{\frac{sin 2 α}{cos 2 α} + \frac{cos α}{sin α}}{\frac{sin 2 α}{cos 2 a l p h a} - \frac{sin α}{cos α}}$ $color(white)(LHS)= ((sin2alpha)/(cos2alpha)+cosalpha/sinalpha)/((sin2alpha)/(cos2alph a)-sinalpha/cosalpha)$

$color(white)(LHS)= ((sin2alphasinalpha+cos2alphacosalpha)/(cancelcos2alphasinalpha ))/((sin2alphacosalpha- cos2alphasinalpha)/(cancelcos2alphacosalpha$

$color(white)(LHS)= (color(red)((cos2alphacosalpha+sin2alphasinalpha))xxcosalpha)/(color(blue)((sin2alphaco salpha-cos2alphasinalpha))xxsinalpha$

Using $(1) and(2)$ , we get

$LHS=(color(red)(cos(2alpha-alpha))xxcosalpha)/(color(blue)(sin(2alpha- alpha))xxsinalpha)$

$color(white)(LHS)=(cosalphaxxcosalpha)/(sinalphaxxsinalpha)$

$color(white)(LHS)=cos^2alpha/sin^2alpha$

$color(white)(LHS)=cot^2alpha$

$LHS=RHS$

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Using $t = tan (\frac{1}{2} θ)$ $t=tan(1/2theta)$ , how do you prove that $\frac{tan 2 α + cot α}{tan 2 α - tan α} = {cot}^{2} α$ $(tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alpha$ ?

2 Answers

Explanation:

Explanation:

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Using t=tan(1/2theta)t=tan(12θ)t=tan(1/2theta), how do you prove that (tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alphatan2α+cotαtan2α−tanα=cot2α(tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alpha?

2 Answers

Explanation:

Explanation:

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Impact of this question

Using $t = tan (\frac{1}{2} θ)$ $t=tan(1/2theta)$ , how do you prove that $\frac{tan 2 α + cot α}{tan 2 α - tan α} = {cot}^{2} α$ $(tan2alpha+cotalpha)/(tan2alpha-tanalpha)=cot^2alpha$ ?