If \(cosec \theta+cot \theta=y\) then establish that \(cos \theta=\frac{y^2-1}{y^2+1}\)
Proof:
\(y=\frac{1+cos \theta}{sin \theta}\)
Therefore, \(y^2=\)
On simplification, \(y^2-1=\)
\(y^2+1=\)
On simplification we get, \(cos \theta=\frac{y^2-1}{y^2+1}\)
Answer variants:
\((cos \theta-1)\)
\(\frac{1+cos^2 \theta+2 cos \theta}{cos^2 \theta}\)
\(2cos \theta(cos \theta+1)\)
\(\frac{1+cos^2 \theta+2 cos \theta}{sin^2 \theta}\)
\(2cos \theta(cos \theta-1)\)
\(2(cos \theta+1)\)