Question

Prove the given trigonometric equation $\dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}} = cosec\theta + \cot \theta $

Answer 1

Hint: Divide the numerator and denominator of LHS by $\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 $\dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}} = cosec\theta + \cot \theta $.
Dividing the LHS by $\sin \theta $, we get-
$LHS = \dfrac{{\cos \theta - \sin \theta + 1}}{{\cos \theta + \sin \theta - 1}} = \dfrac{{\dfrac{1}{{\sin \theta }}(\cos \theta - \sin \theta + 1)}}{{\dfrac{1}{{\sin \theta }}(\cos \theta + \sin \theta - 1)}}$
Further solving we get-
$
  LHS = \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- $\cos e{c^2}A - {\cot ^2}A = 1$ in the LHS we get-
$
  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 ${a^2} - {b^2} = (a + b)(a - b)$}
Taking $(cosec\theta + \cot \theta )$ common from numerator of LHS-
$
  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, $\cos e{c^2}A - {\cot ^2}A = 1$ and algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$ to prove R.H.S = L.H.S.