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Question: Prove the given trigonometric equation \(\dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin...

Prove the given trigonometric equation 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

Hint: Divide the numerator and denominator of LHS by sin⁡θ\sin \theta and then solve the expression and prove. This problem is totally based on trigonometric identities, So however by doing any operation we convert this into standard identities.

Complete step-by-step solution -
We have to 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 .
Dividing the LHS by sin⁡θ\sin \theta , we get-
LHS=cos⁡θ−sin⁡θ+1cos⁡θ+sin⁡θ−1=1sin⁡θ(cos⁡θ−sin⁡θ+1)1sin⁡θ(cos⁡θ+sin⁡θ−1)LHS = \dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}} = \dfrac{{\dfrac{1}{{\sin \theta }}(\cos \theta - \sin \theta + 1)}}{{\dfrac{1}{{\sin \theta }}(\cos \theta + \sin \theta - 1)}}
Further solving we get-
LHS=1sin⁡θ(cos⁡θ−sin⁡θ+1)1sin⁡θ(cos⁡θ+sin⁡θ−1) =(cot⁡θ−1+cosecθ)(cot⁡θ+1−cosecθ) =(cot⁡θ+cosecθ−1)(cot⁡θ−cosecθ+1)  LHS = \dfrac{{\dfrac{1}{{\sin \theta }}(\cos \theta - \sin \theta + 1)}}{{\dfrac{1}{{\sin \theta }}(\cos \theta + \sin \theta - 1)}} \\\ = \dfrac{{(\cot \theta - 1 + cosec\theta )}}{{(\cot \theta + 1 - cosec\theta )}} \\\ = \dfrac{{(\cot \theta + cosec\theta - 1)}}{{(\cot \theta - cosec\theta + 1)}} \\\
Using the trigonometric identity- cos⁡ec2A−cot⁡2A=1\cos e{c^2}A - {\cot ^2}A = 1 in the LHS we get-
LHS=(cot⁡θ+cosecθ)−(cos⁡ec2θ−cot⁡2θ)(cot⁡θ−cosecθ+1) =(cot⁡θ+cosecθ)−(cosecθ−cot⁡θ)(cosecθ+cot⁡θ)(cot⁡θ−cosecθ+1)  LHS = \dfrac{{(\cot \theta + cosec\theta ) - (\cos e{c^2}\theta - {{\cot }^2}\theta )}}{{(\cot \theta - cosec\theta + 1)}} \\\ = \dfrac{{(\cot \theta + cosec\theta ) - (cosec\theta - \cot \theta )(cosec\theta + \cot \theta )}}{{(\cot \theta - cosec\theta + 1)}} \\\
(by algebraic identity a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b)}
Taking (cosecθ+cot⁡θ)(cosec\theta + \cot \theta ) common from numerator of LHS-
LHS=(cot⁡θ+cosecθ)(1−cosecθ+cot⁡θ)(1−cosecθ+cot⁡θ) =cosecθ+cot⁡θ=RHS  LHS = \dfrac{{(\cot \theta + cosec\theta )(1 - cosec\theta + \cot \theta )}}{{(1 - cosec\theta + \cot \theta )}} \\\ = cosec\theta + \cot \theta = RHS \\\
Therefore, LHS = RHS (Hence Proved).

Note: Whenever such types of questions appear, then write down the LHS of the given expression and then divide it by sin(theta) in both numerator and denominator and then solve it by using various trigonometric identities, as mentioned in the solution, cos⁡ec2A−cot⁡2A=1\cos e{c^2}A - {\cot ^2}A = 1 and algebraic identity a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b) to prove R.H.S = L.H.S.