minimopeds
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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!!!
Trig Identities: show 1/sin^2+1/cos^2 equals 1/sin^2-sin^4
- Thread starter minimopeds
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skeeter
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Re: Trig Identities
\(\displaystyle \frac{1}{\sin^2{x}} + \frac{1}{\cos^2{x}} =\)
\(\displaystyle \frac{\cos^2{x}}{\sin^2{x}\cos^2{x}} + \frac{\sin^2{x}}{\sin^2{x}\cos^2{x}} =\)
\(\displaystyle \frac{\cos^2{x} + \sin^2{x}}{\sin^2{x}\cos^2{x}} =\)
\(\displaystyle \frac{1}{\sin^2{x}\cos^2{x}} =\)
\(\displaystyle \frac{1}{\sin^2{x}(1 - \sin^2{x})} =\)
\(\displaystyle \frac{1}{\sin^2{x} - \sin^4{x}}\)
minimopeds said:
\(\displaystyle \frac{1}{\sin^2{x}} + \frac{1}{\cos^2{x}} =\)
\(\displaystyle \frac{\cos^2{x}}{\sin^2{x}\cos^2{x}} + \frac{\sin^2{x}}{\sin^2{x}\cos^2{x}} =\)
\(\displaystyle \frac{\cos^2{x} + \sin^2{x}}{\sin^2{x}\cos^2{x}} =\)
\(\displaystyle \frac{1}{\sin^2{x}\cos^2{x}} =\)
\(\displaystyle \frac{1}{\sin^2{x}(1 - \sin^2{x})} =\)
\(\displaystyle \frac{1}{\sin^2{x} - \sin^4{x}}\)
Re: Trig Identities
\(\displaystyle \text{Hello, minimopeds!}\)
\(\displaystyle \text{Another approach . . .}\)
\(\displaystyle \text{We have: }\;\frac{1}{\sin^2\!x} + \frac{1}{\cos^2\!x} \;=\; \frac{1}{\sin^2\!x} + \frac{1}{1-\sin^2\!x}\)
\(\displaystyle \text{Get a common denominator:}\)
. . \(\displaystyle \frac{1}{\sin^2\!x}\cdot\frac{1-\sin^2\!x}{1-\sin^2\!x} + \frac{1}{1-\sin^2\!x}\cdot\frac{\sin^2\!x}{\sin^2\!x} \;=\;\frac{1-\sin^2\!x + \sin^2\!x}{\sin^2\!x(1-\sin^2\!x)}\)
\(\displaystyle \text{And we have: }\;\frac{1}{\sin^2\!x(1-\sin^2\!x)} \;=\;\frac{1}{\sin^2\!x - \sin^4\!x}\)
\(\displaystyle \text{Hello, minimopeds!}\)
\(\displaystyle \text{Another approach . . .}\)
\(\displaystyle \text{We have: }\;\frac{1}{\sin^2\!x} + \frac{1}{\cos^2\!x} \;=\; \frac{1}{\sin^2\!x} + \frac{1}{1-\sin^2\!x}\)
\(\displaystyle \text{Get a common denominator:}\)
. . \(\displaystyle \frac{1}{\sin^2\!x}\cdot\frac{1-\sin^2\!x}{1-\sin^2\!x} + \frac{1}{1-\sin^2\!x}\cdot\frac{\sin^2\!x}{\sin^2\!x} \;=\;\frac{1-\sin^2\!x + \sin^2\!x}{\sin^2\!x(1-\sin^2\!x)}\)
\(\displaystyle \text{And we have: }\;\frac{1}{\sin^2\!x(1-\sin^2\!x)} \;=\;\frac{1}{\sin^2\!x - \sin^4\!x}\)