Prove that:$\dfrac{{\sec heta - \cos heta + 1}}{{\sin heta + \cos heta - 1}} = \sec heta + an heta $

Question

Prove that:$\dfrac{{\sec 	heta  - \cos 	heta  + 1}}{{\sin 	heta  + \cos 	heta  - 1}} = \sec 	heta  + 	an 	heta $

Vedantu Content Team · Accepted Answer

Hint: Divide the LHS by $\cos 	heta $ and further use trigonometric formulas to convert the lower base into terms of $\sec 	heta $ and $	an 	heta $ so that it could be cancelled out by the numerator. Use rationalizing to get to the answer.Complete step-by-step answer:We know that  ${\sec ^2}	heta  - {	an ^2}	heta  = 1$ LHS –$\dfrac{{\sin 	heta  - \cos 	heta  + 1}}{{\sin 	heta  + \cos 	heta  - 1}}$ Dividing numerator and denominator by $\cos 	heta $ , we get$\dfrac{{	an 	heta  + \sec 	heta  - 1}}{{	an 	heta  - \sec 	heta  + 1}}$ =  $\dfrac{{	an 	heta  + \sec 	heta  - 1}}{{\left( {	an 	heta  - \sec 	heta } ight) + \left( {{{\sec }^2}	heta  - {{	an }^2}	heta } ight)}}$ ( from above)= $\dfrac{{	an 	heta  + \sec 	heta  - 1}}{{ - \left( {\sec 	heta  - 	an 	heta } ight) + \left( {\sec 	heta  - tan	heta } ight)\left( {\sec 	heta  + 	an 	heta } ight)}}$  ( Since ${a^2} - {b^2} = \left( {a + b} ight)\left( {a - b} ight)$ )= $\dfrac{{	an 	heta  + \sec 	heta  - 1}}{{\left( {\sec 	heta  - 	an 	heta } ight)\left( {\sec 	heta  + 	an 	heta  - 1} ight)}}$ = $\dfrac{1}{{\left( {\sec 	heta  - 	an 	heta } ight)}}$ ( cancelling out the common terms)= $\dfrac{1}{{\left( {\sec 	heta  - 	an 	heta } ight)}}\dfrac{{\left( {\sec 	heta  + 	an 	heta } ight)}}{{\left( {\sec 	heta  + 	an 	heta } ight)}}$  ( Rationalizing )= $\dfrac{{\sec 	heta  + 	an 	heta }}{{{{\sec }^2}	heta  - {{	an }^2}	heta }}$ =  $\sec 	heta  + 	an 	heta $ = RHSHence proved.Note: In these questions it is advised to simplify the LHS or the RHS according to their complexity of trigonometric functions .  Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar trigonometric functions to get to the final result .

Prove that:
$\dfrac{{\sec \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \sec \theta + \tan \theta $

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Prove that:$\dfrac{{\sec \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \sec \theta + \tan \theta $