Need help with a Trig Identity: sec^2/tan=(sec)(csc)
- Thread starter willyone
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Count Iblis
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Re: Need help with a Trig Identity
Use: tan = sin/cos; sec=1/cos;csc=1/sin to express everything in terms of cos and sin.
willyone said:
Use: tan = sin/cos; sec=1/cos;csc=1/sin to express everything in terms of cos and sin.
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Your book should have included "x" or "ß" or something, so you'd have the full identity. Without knowing the arguments of the various trig functions, it is difficult to know what is meant to be proven. Sorry.willyone said:
Eliz.
Hello, willyone!
The Count has the best suggestion . . .
\(\displaystyle \L\frac{\sec^2x}{\tan x} \:=\:\frac{\frac{1}{\cos^2x}}{\frac{\sin x}{\cos x}} \;=\;\frac{1}{\cos^2x}\,\cdot\frac{\cos x}{\sin x} \;=\;\frac{1}{\cos x\,\cdot\,\sin x}\;=\;\sec x\,\cdot\,\csc x\)
The Count has the best suggestion . . .
\(\displaystyle \L\frac{\sec^2x}{\tan x} \:=\:\frac{\frac{1}{\cos^2x}}{\frac{\sin x}{\cos x}} \;=\;\frac{1}{\cos^2x}\,\cdot\frac{\cos x}{\sin x} \;=\;\frac{1}{\cos x\,\cdot\,\sin x}\;=\;\sec x\,\cdot\,\csc x\)