Question

Prove that \[\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}\]

Answer 1

Hint: Here R. H. S indicates a term which includes $\sec \theta $ so, we divide the numerator and denominator of L. H. S by $\cos \theta $.

Dividing numerator and denominator by $\cos \theta $ makes the L. H. S term easy. As $\sec \theta $ is reciprocal of $\cos \theta $ so we are dividing L .H .S by $\cos \theta $.
$ \Rightarrow $L. H. S = \[\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} \div \dfrac{{\cos \theta }}{{\cos \theta }}\]

$ \Rightarrow $L. H. S =$\dfrac{{\dfrac{{\sin \theta - \cos \theta + 1}}{{\cos \theta }}}}{{\dfrac{{\sin \theta + \cos \theta - 1}}{{\cos \theta }}}}$ ……. (1)
Now, by using trigonometric identities we will simplify the L. H. S term.
We know that
$\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $, $\dfrac{{\cos \theta }}{{\cos \theta }} = 1$, $\dfrac{1}{{\cos \theta }} = \sec \theta $
Now, putting these values in equation (1), we get
$ \Rightarrow $L. H. S = $\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}$ ……. (2)
Now, using the identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$, equation (2) becomes simple. We replace the value of 1 only in the denominator to get the desired result.
Replacing value of 1 in denominator, we get
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta - \sec \theta + ({{\sec }^2}\theta - {{\tan }^2}\theta )}}$
We know that $({a^2} - {b^2}) = (a - b)(a + b)$, using this property L. H. S can be written as,
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta - \sec \theta + (\sec \theta - \tan \theta )(\sec \theta + \tan \theta )}}$
Taking $(\sec \theta - \tan \theta )$ common from denominator,
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{(\sec \theta - \tan \theta )( - 1 + \sec \theta + \tan \theta )}}$
Now, eliminating the same terms from the numerator and the denominator, we get
$ \Rightarrow $L. H. S =$\dfrac{1}{{\sec \theta - \tan \theta }}$= R. H. S
Hence proved.

Note: To solve such problems one should have the knowledge of trigonometric identities like ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , $se{c^2}\theta - {\tan ^2}\theta = 1$ and $\cos e{c^2}\theta - {\cot ^2}\theta = 1$ which are very useful in solving the problems. Another key point is that we have to see the R. H. S of the question to decide which identity we can use to solve the problem.