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