Question

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

Answer 1

Hint: We solve this question by using the concept of rationalization. We multiply both numerator and denominator by \[\tan \theta + (\sec \theta - 1)\] and then substitute the value of \[1 + {\tan ^2}\theta = {\sec ^2}\theta \] in the numerator and \[{\tan ^2}\theta = {\sec ^2}\theta - 1\] in the denominator and solve. In the end we try to make up factors between numerator and denominator which cancel out to give RHS of the equation.

Complete step-by-step answer:
Solve LHS by grouping the numerator and denominator in such a way that they form a fraction like \[\dfrac{{a + b}}{{a - b}}\]
We write \[\dfrac{{\sec \theta + \tan \theta - 1}}{{\tan \theta - \sec \theta + 1}} = \dfrac{{\tan \theta + (\sec \theta - 1)}}{{\tan \theta - (\sec \theta - 1)}}\]
Now rationalizing the LHS of the equation by multiplying both numerator and denominator by \[\tan \theta + (\sec \theta - 1)\] we get
\[ \Rightarrow \dfrac{{\tan \theta + (\sec \theta - 1)}}{{\tan \theta - (\sec \theta - 1)}} \times \dfrac{{\tan \theta + (\sec \theta - 1)}}{{\tan \theta + (\sec \theta - 1)}} = \dfrac{{\tan \theta + (\sec \theta - 1) \times \tan \theta + (\sec \theta - 1)}}{{\tan \theta - (\sec \theta - 1) \times \tan \theta + (\sec \theta - 1)}}\]
Since we know \[(a + b)(a - b) = {a^2} - {b^2}\]
Therefore we can write
\[ \Rightarrow \dfrac{{{{\left[ {\tan \theta + (\sec \theta - 1)} \right]}^2}}}{{{{\tan }^2}\theta - {{(\sec \theta - 1)}^2}}}\]
Now opening up square using the property \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]
\[ \Rightarrow \dfrac{{{{\tan }^2}\theta + ({{\sec }^2}\theta + 1 - 2\sec \theta ) + 2\tan \theta (\sec \theta - 1)}}{{{{\tan }^2}\theta - ({{\sec }^2}\theta + 1 - 2\sec \theta )}}\]
Multiplying the values in the bracket.
\[ \Rightarrow \dfrac{{{{\tan }^2}\theta + {{\sec }^2}\theta + 1 - 2\sec \theta + 2\tan \theta \sec \theta - 2\tan \theta }}{{{{\tan }^2}\theta - {{\sec }^2}\theta - 1 + 2\sec \theta }}\]
\[ \Rightarrow \dfrac{{1 + {{\tan }^2}\theta + {{\sec }^2}\theta - 2\sec \theta + 2\tan \theta \sec \theta - 2\tan \theta }}{{{{\tan }^2}\theta - {{\sec }^2}\theta - 1 + 2\sec \theta }}\]
Substitute the value of \[1 + {\tan ^2}\theta = {\sec ^2}\theta \] in the numerator and \[{\tan ^2}\theta = {\sec ^2}\theta - 1\] in the denominator and solve.
\[ \Rightarrow \dfrac{{{{\sec }^2}\theta + {{\sec }^2}\theta - 2\sec \theta + 2\tan \theta \sec \theta - 2\tan \theta }}{{{{\sec }^2}\theta - 1 - {{\sec }^2}\theta - 1 + 2\sec \theta }}\]
Adding the like terms
\[
   \Rightarrow \dfrac{{2{{\sec }^2}\theta - 2\sec \theta + 2\tan \theta \sec \theta - 2\tan \theta }}{{{{\sec }^2}\theta - {{\sec }^2}\theta - 2 + 2\sec \theta }} \\
   \Rightarrow \dfrac{{2{{\sec }^2}\theta - 2\sec \theta + 2\tan \theta \sec \theta - 2\tan \theta }}{{2\sec \theta - 2}} \\
 \]
Now take 2 common in both numerator and denominator.
\[ \Rightarrow \dfrac{{2({{\sec }^2}\theta - \sec \theta + \tan \theta \sec \theta - \tan \theta )}}{{2(\sec \theta - 1)}}\]
Cancel out the same factor from both numerator and denominator
\[ \Rightarrow \dfrac{{({{\sec }^2}\theta - \sec \theta + \tan \theta \sec \theta - \tan \theta )}}{{(\sec \theta - 1)}}\]
Now we divide the numerator into two parts so when we take a value common we get another value common same as the denominator.
\[ \Rightarrow \dfrac{{({{\sec }^2}\theta - \sec \theta ) + (\tan \theta \sec \theta - \tan \theta )}}{{(\sec \theta - 1)}}\]
Now we take \[\sec \theta \] common from the first bracket and \[\tan \]common from the second bracket
\[ \Rightarrow \dfrac{{\sec \theta (\sec \theta - 1) + \tan \theta (\sec \theta - 1)}}{{(\sec \theta - 1)}}\]
Taking \[(\sec \theta - 1)\]common in the denominator we can write
\[ \Rightarrow \dfrac{{(\sec \theta + \tan \theta )(\sec \theta - 1)}}{{(\sec \theta - 1)}}\]
Cancel out same terms from numerator and denominator
\[ \Rightarrow \sec \theta + \tan \theta \]
=RHS of the equation.
Hence Proved

Note: Students should always keep in mind the sign outside the bracket, because it gets multiplied with all the values inside the bracket so all the values inside the bracket change according to it, students usually make mistakes while opening the brackets with negative signs with them.