solve this proof: sin(x)^2 cot(x)^2= (1-sinx)(1+sinx)
- Thread starter bballbink
- Start date
Re: solve this proof
Hello, bballbink!
If you know any of the basic identities, it's quite easy.
\(\displaystyle \text{We know that: }\:\cot x \:=\:\frac{\cos x}{\sin x}\)
\(\displaystyle \text{So the left side is: }\:\sin^2\!x\cot^2\!x \:=\:\sin^2\!x\cdot\frac{\cos^2\!x}{\sin^2\!x}\)
. . \(\displaystyle \text{which reduces to: }\;\cos^2\!x\)
. . \(\displaystyle \text{which is equal to: }\:1 - \sin^2\!x\)
. . \(\displaystyle \text{which factors: }\;(1 - \sin x)(1 + \sin x)\)
Hello, bballbink!
If you know any of the basic identities, it's quite easy.
\(\displaystyle \sin^2\!x\cot^2\!x \:=\1-\sin x)(1+\sin x)\)
\(\displaystyle \text{We know that: }\:\cot x \:=\:\frac{\cos x}{\sin x}\)
\(\displaystyle \text{So the left side is: }\:\sin^2\!x\cot^2\!x \:=\:\sin^2\!x\cdot\frac{\cos^2\!x}{\sin^2\!x}\)
. . \(\displaystyle \text{which reduces to: }\;\cos^2\!x\)
. . \(\displaystyle \text{which is equal to: }\:1 - \sin^2\!x\)
. . \(\displaystyle \text{which factors: }\;(1 - \sin x)(1 + \sin x)\)