Verify the following identity...Need help

**Brndo4u** · January 25th 2010, 07:58 PM

Use x,y and r definitions of the Trig functions to verify the following identity.

Tan^2 theta Csc theta+1/sin theta= Csc theta Sec^2 theta

**VonNemo19** · January 25th 2010, 08:33 PM

Originally Posted by Brndo4u

Use x,y and r definitions of the Trig functions to verify the following identity.

Tan^2 theta Csc theta+1/sin theta= Csc theta Sec^2 theta

So start by rewriting the left hand side,

$\overbrace{\left(\frac{y}{x}\right)^2}^{\tan^2\the ta}\overbrace{\cdot\frac{r}{y}}^{\csc\theta}+\over brace{\frac{r}{y}}^{\frac{1}{\sin\theta}}$ .

Do the right hand side in a similar fashion, and then manipulate one of the sides so that it looks lik the other if necessary.

**Soroban** · January 25th 2010, 10:26 PM

Hello, Brndo4u!

Use x,y and r definitions of the Trig functions to verify the following identity.

. . $\tan^2\theta\csc\theta + \frac{1}{\sin\theta} \:=\: \csc\theta\sec^2\theta$

The right side is: . $\csc\theta\sec^2\theta \:=\:\left(\frac{r}{y}\right)\left(\frac{r}{x}\rig ht)^2 \:=\:\frac{r^3}{x^2y}$

The left side is: . $\tan^2\theta + \csc\theta + \frac{1}{\sin\theta} \;\;=\;\;\left(\frac{y}{x}\right)^2\left(\frac{r}{ y}\right) + \frac{1}{\left(\frac{y}{r}\right)}$

. . . $=\;\;\frac{ry^2}{x^2y} + \frac{r}{y} \;=\;\frac{ry^2}{x^2y} + \frac{rx^2}{x^2y} \;\;=\;\;\frac{ry^2 + rx^2}{x^2y}$

. . . $=\;\;\frac{r\overbrace{(x^2+y^2)}^{\text{This is }r^2}}{x^2y} \;\;=\;\;\frac{r\cdot r^2}{x^2y} \;\;=\;\;\frac{r^3}{x^2y}$

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