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# Prove that $\dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}} = \ cosec\theta + \cot \theta$

Last updated date: 13th Jun 2024
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Hint: We solve this question by grouping together the term $(1 - \sin \theta )$ from both numerator and denominator and then rationalizing the term by multiplying both numerator and denominator by the same value. Using the trigonometric identities like ${\cos ^2}\theta + {\sin ^2}\theta = 1$ we solve the LHS.

Consider the Left hand side of the equation
$\dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}}$
Group together the term $(1 - \sin \theta )$ from both numerator and denominator
$\Rightarrow \dfrac{{\cos \theta + (1 - \sin \theta )}}{{\cos \theta - (1 - \sin \theta )}}$
Rationalize the fraction by multiplying both numerator and denominator by $\cos \theta + (1 - \sin \theta )$.
$\Rightarrow \dfrac{{\cos \theta + (1 - \sin \theta )}}{{\cos \theta - (1 - \sin \theta )}} \times \dfrac{{\cos \theta + (1 - \sin \theta )}}{{\cos \theta + (1 - \sin \theta )}} \\ \Rightarrow \dfrac{{{{\left( {\cos \theta + (1 - \sin \theta )} \right)}^2}}}{{\left( {\cos \theta - (1 - \sin \theta )} \right) \times \left( {\cos \theta - (1 - \sin \theta )} \right)}} \\$
Using the property $(a + b)(a - b) = {a^2} - {b^2}$, where $a = \cos \theta ,b = (1 - \sin \theta )$
$\Rightarrow \dfrac{{{{\left( {\cos \theta + (1 - \sin \theta )} \right)}^2}}}{{\left( {{{\cos }^2}\theta - {{(1 - \sin \theta )}^2}} \right)}}$
Now opening the squares using the property ${(a - b)^2} = {a^2} + {b^2} - 2ab$ in denominator and the property ${(a + b)^2} = {a^2} + {b^2} + 2ab$ in the numerator.
$\Rightarrow \dfrac{{\left( {{{\cos }^2}\theta + {{(1 - \sin \theta )}^2} + 2\cos \theta (1 - \sin \theta )} \right)}}{{\left( {{{\cos }^2}\theta - (1 + {{\sin }^2}\theta - 2\sin \theta )} \right)}} \\ \Rightarrow \dfrac{{\left( {{{\cos }^2}\theta + (1 + {{\sin }^2}\theta - 2\sin \theta ) + 2\cos \theta - 2\cos \theta \sin \theta } \right)}}{{\left( {{{\cos }^2}\theta - 1 - {{\sin }^2}\theta + 2\sin \theta } \right)}} \\ \Rightarrow \dfrac{{\left( {{{\cos }^2}\theta + 1 + {{\sin }^2}\theta - 2\sin \theta + 2\cos \theta - 2\cos \theta \sin \theta } \right)}}{{\left( {{{\cos }^2}\theta - 1 - {{\sin }^2}\theta + 2\sin \theta } \right)}} \\$
Now we pair up the terms that can be transformed using trigonometric identities.
$\Rightarrow \dfrac{{\left( {({{\cos }^2}\theta + {{\sin }^2}\theta ) + 1 - 2\sin \theta + 2\cos \theta - 2\cos \theta \sin \theta } \right)}}{{\left( {{{\cos }^2}\theta - 1 - {{\sin }^2}\theta + 2\sin \theta } \right)}}$
We know that ${\cos ^2}\theta + {\sin ^2}\theta = 1 \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta$
We substitute ${\cos ^2}\theta + {\sin ^2}\theta = 1$ in the numerator and ${\cos ^2}\theta = 1 - {\sin ^2}\theta$ in the denominator.
$\Rightarrow \dfrac{{\left( {1 + 1 - 2\sin \theta + 2\cos \theta - 2\cos \theta \sin \theta } \right)}}{{\left( {1 - {{\sin }^2}\theta - 1 - {{\sin }^2}\theta + 2\sin \theta } \right)}} \\ \Rightarrow \dfrac{{\left( {2 - 2\sin \theta + 2\cos \theta - 2\cos \theta \sin \theta } \right)}}{{\left( { - 2{{\sin }^2}\theta + 2\sin \theta } \right)}} \\$
Now we take 2 common from both denominator and numerator.
$\Rightarrow \dfrac{{2\left( {1 - \sin \theta + \cos \theta - \cos \theta \sin \theta } \right)}}{{2\left( {\sin \theta - {{\sin }^2}\theta } \right)}}$
Cancel the same terms from both numerator and denominator.
$\Rightarrow \dfrac{{\left( {1 - \sin \theta + \cos \theta - \cos \theta \sin \theta } \right)}}{{\left( {\sin \theta - {{\sin }^2}\theta } \right)}}$
Now we take the terms common in the denominator and numerator.
$\Rightarrow \dfrac{{\left( {1 - \sin \theta ) + \cos \theta (1 - \sin \theta } \right)}}{{\sin \theta \left( {1 - \sin \theta } \right)}} \\ \Rightarrow \dfrac{{(1 - \sin \theta )(1 + \cos \theta )}}{{\sin \theta \left( {1 - \sin \theta } \right)}} \\$
Cancel out the same terms from both numerator and denominator.
$\Rightarrow \dfrac{{1 + \cos \theta }}{{\sin \theta }}$
Now break the fraction into two parts
$\Rightarrow \dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}$
Since, we know that $\ cosec\theta = \dfrac{1}{{\sin \theta }},\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$, so substitute the values in the equation
$\Rightarrow \cos ec\theta + \cot \theta$
which is equal to RHS of the equation.
Hence Proved

Note: Students many times make mistake of grouping wrong terms in the starting of the solution, always keep in mind that we have to create numerator of the type $(a + b)$ and denominator of the type $(a - b)$ or vice versa so when we rationalize the term in the numerator gets squared and the term in the denominator becomes easy so we can apply the formula $(a + b)(a - b) = {a^2} - {b^2}$ to it.
Also, many students group together $2\sin \theta \cos \theta = \sin 2\theta$ which should not be done because then we will not be able to cancel out common factor 2 from numerator and denominator, which will make our solution complex.