Question

Find \(k\) if
\[
\begin{array}{r}
1-\frac{\sin ^{2} \alpha}{1+\cot \alpha}-\frac{\cos ^{2} \alpha}{1+\tan \alpha} \\
=k \sin \alpha \cos \alpha
\end{array}
\]

Answer 1

\(1-\frac{\sin ^{2} \alpha}{1+\cot \alpha}-\frac{\cos ^{2} \alpha}{1+\tan \alpha}=k \sin \alpha \cos \alpha\)

\(1-\frac{\sin ^{2} \alpha}{1+\frac{\cos \alpha}{\sin \alpha}}-\frac{\cos ^{2}}{1+\frac{\sin \alpha}{\cos \alpha}}=k \sin \alpha \cos \alpha\)

\(1-\frac{\sin ^{3} \alpha}{\sin \alpha+\cos \alpha}-\frac{\cos ^{3} \alpha}{\sin \alpha+\cos \alpha}=k \sin \alpha \cos \alpha\)

\(\frac{\sin \alpha+\cos \alpha-\sin ^{3} \alpha-\cos ^{3} \alpha}{\sin \alpha+\cos \alpha}=k \sin \alpha \cos \alpha\)

\(\frac{\sin \alpha\left(1-\sin ^{2} \alpha\right)+\cos \alpha\left(1-\cos ^{2} \alpha\right)}{\sin \alpha+\cos \alpha}=k \sin \alpha \cos \alpha\)

\(\frac{\sin \alpha \cos 2 \alpha+\cos \alpha \sin 2 \alpha}{\sin \alpha+\cos \alpha}=k \sin \alpha \cos \alpha\)

\(\frac{\sin \alpha \cos \alpha(\sin \alpha+\cos \alpha)}{\sin \alpha+\cos \alpha}=k \sin \alpha \cos \alpha\)

\(\frac{\sin \alpha \cos \alpha}{\sin \alpha \cos \alpha} \times \frac{\sin \alpha+\cos \alpha}{\sin \alpha+\cos \alpha}=k\)
\(k=1\)