Question

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

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Hint: Divide the LHS by $\cos \theta$ and further use trigonometric formulas to convert the lower base into terms of $\sec \theta$ and $\tan \theta$ so that it could be cancelled out by the numerator. Use rationalizing to get to the answer.

We know that ${\sec ^2}\theta - {\tan ^2}\theta = 1$
LHS –
$\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}}$
Dividing numerator and denominator by $\cos \theta$ , we get
$\dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}}$
= $\dfrac{{\tan \theta + \sec \theta - 1}}{{\left( {\tan \theta - \sec \theta } \right) + \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}$ ( from above)
= $\dfrac{{\tan \theta + \sec \theta - 1}}{{ - \left( {\sec \theta - \tan \theta } \right) + \left( {\sec \theta - tan\theta } \right)\left( {\sec \theta + \tan \theta } \right)}}$ ( Since ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ )
= $\dfrac{{\tan \theta + \sec \theta - 1}}{{\left( {\sec \theta - \tan \theta } \right)\left( {\sec \theta + \tan \theta - 1} \right)}}$
= $\dfrac{1}{{\left( {\sec \theta - \tan \theta } \right)}}$ ( cancelling out the common terms)
= $\dfrac{1}{{\left( {\sec \theta - \tan \theta } \right)}}\dfrac{{\left( {\sec \theta + \tan \theta } \right)}}{{\left( {\sec \theta + \tan \theta } \right)}}$ ( Rationalizing )
= $\dfrac{{\sec \theta + \tan \theta }}{{{{\sec }^2}\theta - {{\tan }^2}\theta }}$ = $\sec \theta + \tan \theta$ = RHS
Hence 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 .