sin2θ tanθ + cos2θ cot θ + 2sin θ cos θ = tan θ + cot θ

Question:

sin2θ tanθ + cos2θ cot θ + 2sin θ cos θ = tan θ + cot θ

Solution:

$\sin ^{2} \theta \tan \theta+\cos ^{2} \theta \cot \theta+2 \sin \theta \cos \theta$

$=\frac{\sin ^{3} \theta}{\cos \theta}+\frac{\cos ^{3} \theta}{\sin \theta}+2 \sin \theta \cos \theta$

$=\frac{\sin ^{4} \theta+\cos ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta}{\sin \theta \cos \theta}$

$=\frac{\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}}{\sin \theta \cos \theta}$

$=\frac{1}{\sin \theta \cos \theta} \quad\left(\sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}$

$=\frac{\sin ^{2} \theta}{\sin \theta \cos \theta}+\frac{\cos ^{2} \theta}{\sin \theta \cos \theta}$

$=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$

$=\tan \theta+\cot \theta$

 

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