Question

Show that $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\csc \theta$

Answer 1

Let usfirstfind the value of left hand side (LHS) that is $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }$ as shown below:

$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=\frac{(\sin \theta \times \sin \theta )+(1+\cos \theta )(1+\cos \theta )}{\sin \theta (1+\cos \theta )}=\frac{{\sin }^2\theta +(1+\cos \theta {)}^2}{\sin \theta (1+\cos \theta )}=\frac{{\sin }^2\theta +(1+{\cos }^2\theta +2\cos \theta )}{\sin \theta (1+\cos \theta )}(\because (a+b{)}^2=a^2+b^2+2ab)=\frac{{\sin }^2\theta +{\cos }^2\theta +1+2\cos \theta }{\sin \theta (1+\cos \theta )}$

$=\frac{1+1+2\cos \theta }{\sin \theta (1+\cos \theta )}(\because {\sin }^{2}x+{\cos }^{2}x=1)=\frac{2+2\cos \theta }{\sin \theta (1+\cos \theta )}=\frac{2(1+\cos \theta )}{\sin \theta (1+\cos \theta )}=\frac{2}{\sin \theta }=2\csc \theta =RHS(\because \csc x=\frac{1}{\sin x})$

Since LHS=RHS,

Hence, $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\csc \theta$.