Question

Prove the following trigonometric identities.

(i) ${\left(\frac{1+\sin \theta -\cos \theta }{1+\sin \theta +\cos \theta }\right)}^2=\frac{1-\cos \theta }{1+\cos \theta }$
(ii) $\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }=\frac{1-\sin \theta }{\cos {\displaystyle \theta }}$

Answer 1

(i) In the given question, we need to prove

Taking common from the numerator and the denominator of the L.H.S, we get

Now, using the property , we get

Using , we get

Taking common from the numerator, we get

Using and , we get

Now, using the property , we get

Hence proved.

(ii)
Consider the LHS.
$$\begin{gathered} \frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta } \\ =\frac{{\displaystyle \left(\sec \theta -\tan \theta \right)+1}}{{\displaystyle 1+\sec \theta +\tan \theta }} \\ =\frac{{\displaystyle \left(\sec \theta -\tan \theta \right)+\left({\sec }^2\theta -{\tan }^2\theta \right)}}{{\displaystyle 1+\sec \theta +\tan \theta }}\left({\sec }^2\theta -{\tan }^2\theta =1\right) \\ =\frac{{\displaystyle \left(\sec \theta -\tan \theta \right)\left(1+\sec \theta +\tan \theta \right)}}{{\displaystyle 1+\sec \theta +\tan \theta }} \\ =\sec \theta -\tan \theta \\ =\frac{1}{\cos {\displaystyle \theta }}-\frac{{\displaystyle \sin \theta }}{{\displaystyle \cos \theta }} \\ =\frac{{\displaystyle 1-\sin \theta }}{{\displaystyle \cos \theta }} \end{gathered}$$
= RHS
Hence proved.