If \(cosec \ \beta+ cot \ \beta = y\), then show that \(cos \ \beta= \frac{y^2 - 1}{y^2 + 1}\).
Proof:
\(cosec \ \beta + cot \ \beta = y\)
Squaring on both sides, we get:
\(1+ cos \ \beta= y^2 (1-cos\ \beta)\)
\(1+ cos \ \beta= y^2 - y^2cos\ \beta\)
\(cos \ \beta+ y^2cos\ \beta=y^2-1\)
\(cos\ \beta(1+y^2)=y^2-1\)
\(cos\ \beta=\frac{y^2-1}{y^2+1}\)
Hence, we proved.
Answer variants:
\(\frac{1+cos \ \beta}{sin\ \beta}=y\)
\(\frac{(1+cos\ \beta)^2}{1-cos^2\ \beta}=y^{2}\)
\(\frac{(1+cos\ \beta)^2}{(1-cos\ \beta)(1+cos\ \beta)}=y^{2}\)
\(\frac{1}{sin \ \beta}+\frac{cos\ \beta}{sin \ \beta}=y\)
\(\frac{1+cos\ \beta}{1-cos\ \beta}=y^2\)