Trig Identities: show 1/sin^2+1/cos^2 equals 1/sin^2-sin^4

[minimopeds]

How do you show that 1/sin^2+1/cos^2 is equal to 1/sin^2-sin^4. Please Do give me some help!!!

 

[skeeter]

Re: Trig Identities

How do you show that 1/sin^2+1/cos^2 is equal to 1/sin^2-sin^4. Please Do give me some help!!!

$\displaystyle \frac{1}{\sin^2{x}} + \frac{1}{\cos^2{x}} =$

$\displaystyle \frac{\cos^2{x}}{\sin^2{x}\cos^2{x}} + \frac{\sin^2{x}}{\sin^2{x}\cos^2{x}} =$

$\displaystyle \frac{\cos^2{x} + \sin^2{x}}{\sin^2{x}\cos^2{x}} =$

$\displaystyle \frac{1}{\sin^2{x}\cos^2{x}} =$

$\displaystyle \frac{1}{\sin^2{x}(1 - \sin^2{x})} =$

$\displaystyle \frac{1}{\sin^2{x} - \sin^4{x}}$

 

[soroban]

Re: Trig Identities

$\displaystyle \text{Hello, minimopeds!}$


$\displaystyle \text{Another approach . . .}$

$\displaystyle \boxed{\text{Show that: }\;\frac{1}{\sin^2x} +\frac{1}{\cos^2x} \;=\;\frac{1}{\sin^2\!x-\sin^4\!x}}$

$\displaystyle \text{We have: }\;\frac{1}{\sin^2\!x} + \frac{1}{\cos^2\!x} \;=\; \frac{1}{\sin^2\!x} + \frac{1}{1-\sin^2\!x}$


$\displaystyle \text{Get a common denominator:}$

. . $\displaystyle \frac{1}{\sin^2\!x}\cdot\frac{1-\sin^2\!x}{1-\sin^2\!x} + \frac{1}{1-\sin^2\!x}\cdot\frac{\sin^2\!x}{\sin^2\!x} \;=\;\frac{1-\sin^2\!x + \sin^2\!x}{\sin^2\!x(1-\sin^2\!x)}$


$\displaystyle \text{And we have: }\;\frac{1}{\sin^2\!x(1-\sin^2\!x)} \;=\;\frac{1}{\sin^2\!x - \sin^4\!x}$