Question

Prove that $\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}$ using the identity ${\sec ^2}\theta = 1 + {\tan ^2}\theta$.

Answer 1

We use the value of trigonometric functions like tangent and secant in terms of sine and cosine to prove the LHS of the equation equal to RHS of the equation. Divide both numerator and denominator by $\cos \theta$ to convert the equation in form of tangent and secant. Multiply the fraction with a fraction that has a numerator and denominator as the same, so we can make use of the given identity.

• Value of $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
• Value of $\sec \theta = \dfrac{1}{{\cos \theta }}$

Complete step-by-step answer:
We have to prove $\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}$
We solve the left hand side of the equation
LHS of the equation is $\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}}$
Divide both numerator and denominator by$\cos \theta$.
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{\dfrac{{\sin \theta - \cos \theta + 1}}{{\cos \theta }}}}{{\dfrac{{\sin \theta + \cos \theta - 1}}{{\cos \theta }}}}$
Separate the division in fraction
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }} - \dfrac{{\cos \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }}}}{{\dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\cos \theta }} - \dfrac{1}{{\cos \theta }}}}$
Since we know the value of$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$and value of $\sec \theta = \dfrac{1}{{\cos \theta }}$
Substitute the values in numerator and denominator.
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}$
Now multiply both numerator and denominator by same factor i.e. $(\tan \theta - \sec \theta )$
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }} \times \dfrac{{(\tan \theta - \sec \theta )}}{{(\tan \theta - \sec \theta )}}$
Multiply the values in numerator only$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{{{\tan }^2}\theta - \tan \theta + \sec \theta \tan \theta - \sec \theta \tan \theta + \sec \theta - {{\sec }^2}\theta }}{{(\tan \theta + 1 - \sec \theta )(\tan \theta - \sec \theta )}}$
Cancel the terms having same magnitude but opposite signs
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{{{\tan }^2}\theta - \tan \theta + \sec \theta - {{\sec }^2}\theta }}{{(\tan \theta + 1 - \sec \theta )(\tan \theta - \sec \theta )}}$
Substitute the value of ${\sec ^2}\theta = 1 + {\tan ^2}\theta$
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{{{\tan }^2}\theta - \tan \theta + \sec \theta - 1 - {{\tan }^2}\theta }}{{(\tan \theta + 1 - \sec \theta )(\tan \theta - \sec \theta )}}$
Cancel the terms having same magnitude but opposite signs
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{ - 1 - \tan \theta + \sec \theta }}{{(\tan \theta + 1 - \sec \theta )(\tan \theta - \sec \theta )}}$
Take negative sign common from numerator
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{ - (\tan \theta + 1 - \sec \theta )}}{{(\tan \theta + 1 - \sec \theta )(\tan \theta - \sec \theta )}}$
Cancel the same factors from numerator and denominator.
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{ - 1}}{{(\tan \theta - \sec \theta )}}$
Multiply both numerator and denominator by -1
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{{ - 1}}{{(\tan \theta - \sec \theta )}} \times \dfrac{{ - 1}}{{ - 1}}$
Since we know the multiplication of two negative signs gives a positive sign
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{( - \tan \theta + \sec \theta )}}$
$\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}$
RHS of the equation is $\dfrac{1}{{\sec \theta - \tan \theta }}$
$\therefore$LHS $=$RHS
Hence proved

Note: Students might make the mistake of proving this question by rationalizing the process. Keep in mind we rationalize the fraction where we have to find the smallest possible value of the function, if we rationalize the LHS we will have to rationalize the RHS as well. Here we take the hint of which value to be multiplied in numerator and denominator by looking at the RHS.
$\cos ec \theta = \dfrac{1}{{\sin \theta }}$
$\tan \theta = \dfrac{1}{{\cot \theta }}$