Question

Show that $\frac{1+\sin \theta }{1-\sin \theta }={(\sec \theta +\tan \theta )}^2$

Answer 1

Let usfirstfind the value of left hand side (LHS) that is $\frac{1+\sin \theta }{1-\sin \theta }$ by multiplying and dividing it by $\left(1+\sin \theta \right)$ as shown below:

$\frac{1+\sin \theta }{1-\sin \theta }\times \frac{1+\sin \theta }{1+\sin \theta }=\frac{(1+\sin \theta {)}^2}{1-{\sin }^2\theta }(\because (a-b\left)\right(a+b)=a^2-b^2)=\frac{(1+\sin \theta {)}^2}{{\cos }^2\theta }(\because 1-{\sin }^{2}x={\cos }^{2}x)={\left(\frac{1+\sin \theta }{\cos \theta }\right)}^2={(\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta })}^2={(\sec \theta +\tan \theta )}^2=RHS(\because \sec x=\frac{1}{\cos x},\tan x=\frac{\sin x}{\cos x})$

Since LHS=RHS,

Hence, $\frac{1+\sin \theta }{1-\sin \theta }={(\sec \theta +\tan \theta )}^2$.