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Question: Prove that \[\dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}} = \ cosec\thet...

Prove that cos⁡θ−sin⁡θ+1cos⁡θ+sin⁡θ−1= cosecθ+cot⁡θ\dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}} = \ cosec\theta + \cot \theta

Explanation

Solution

We solve this question by grouping together the term (1−sin⁡θ)(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⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 we solve the LHS.

Complete step-by-step answer:
Consider the Left hand side of the equation
cos⁡θ−sin⁡θ+1cos⁡θ+sin⁡θ−1\dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}}
Group together the term (1−sin⁡θ)(1 - \sin \theta ) from both numerator and denominator
⇒cos⁡θ+(1−sin⁡θ)cos⁡θ−(1−sin⁡θ)\Rightarrow \dfrac{{\cos \theta + (1 - \sin \theta )}}{{\cos \theta - (1 - \sin \theta )}}
Rationalize the fraction by multiplying both numerator and denominator by cos⁡θ+(1−sin⁡θ)\cos \theta + (1 - \sin \theta ).

⇒cos⁡θ+(1−sin⁡θ)cos⁡θ−(1−sin⁡θ)×cos⁡θ+(1−sin⁡θ)cos⁡θ+(1−sin⁡θ) ⇒(cos⁡θ+(1−sin⁡θ))2(cos⁡θ−(1−sin⁡θ))×(cos⁡θ−(1−sin⁡θ))  \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)=a2−b2(a + b)(a - b) = {a^2} - {b^2}, where a=cos⁡θ,b=(1−sin⁡θ)a = \cos \theta ,b = (1 - \sin \theta )
⇒(cos⁡θ+(1−sin⁡θ))2(cos⁡2θ−(1−sin⁡θ)2)\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=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab in denominator and the property (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab in the numerator.

⇒(cos⁡2θ+(1−sin⁡θ)2+2cos⁡θ(1−sin⁡θ))(cos⁡2θ−(1+sin⁡2θ−2sin⁡θ)) ⇒(cos⁡2θ+(1+sin⁡2θ−2sin⁡θ)+2cos⁡θ−2cos⁡θsin⁡θ)(cos⁡2θ−1−sin⁡2θ+2sin⁡θ) ⇒(cos⁡2θ+1+sin⁡2θ−2sin⁡θ+2cos⁡θ−2cos⁡θsin⁡θ)(cos⁡2θ−1−sin⁡2θ+2sin⁡θ)  \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.
⇒((cos⁡2θ+sin⁡2θ)+1−2sin⁡θ+2cos⁡θ−2cos⁡θsin⁡θ)(cos⁡2θ−1−sin⁡2θ+2sin⁡θ)\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⁡2θ+sin⁡2θ=1⇒cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta + {\sin ^2}\theta = 1 \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta
We substitute cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 in the numerator and cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta in the denominator.

⇒(1+1−2sin⁡θ+2cos⁡θ−2cos⁡θsin⁡θ)(1−sin⁡2θ−1−sin⁡2θ+2sin⁡θ) ⇒(2−2sin⁡θ+2cos⁡θ−2cos⁡θsin⁡θ)(−2sin⁡2θ+2sin⁡θ)  \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.
⇒2(1−sin⁡θ+cos⁡θ−cos⁡θsin⁡θ)2(sin⁡θ−sin⁡2θ)\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.
⇒(1−sin⁡θ+cos⁡θ−cos⁡θsin⁡θ)(sin⁡θ−sin⁡2θ)\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.

⇒(1−sin⁡θ)+cos⁡θ(1−sin⁡θ)sin⁡θ(1−sin⁡θ) ⇒(1−sin⁡θ)(1+cos⁡θ)sin⁡θ(1−sin⁡θ)  \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.
⇒1+cos⁡θsin⁡θ\Rightarrow \dfrac{{1 + \cos \theta }}{{\sin \theta }}
Now break the fraction into two parts
⇒1sin⁡θ+cos⁡θsin⁡θ\Rightarrow \dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}
Since, we know that  cosecθ=1sin⁡θ,cot⁡θ=cos⁡θsin⁡θ\ cosec\theta = \dfrac{1}{{\sin \theta }},\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}, so substitute the values in the equation
⇒cos⁡ecθ+cot⁡θ\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)(a + b) and denominator of the type (a−b)(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)=a2−b2(a + b)(a - b) = {a^2} - {b^2} to it.
Also, many students group together 2sin⁡θcos⁡θ=sin⁡2θ2\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.