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Question: Prove the following trigonometric equation: \[\dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \thet...

Prove the following trigonometric equation:
tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1=1+sin⁡θcos⁡θ\dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}} = \dfrac{{1 + \sin \theta }}{{\cos \theta }}

Explanation

Solution

Hint: To prove this question we have to start from LHS and using standard results like (sec⁡2θ−tan⁡2θ=1)\left( {{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right) we have to proceed to get RHS. To get RHS we have to use some common sense that what changes should be made to proceed further.

Complete step-by-step answer:
We have LHS,
⇒tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1\Rightarrow \dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}}
Here we are not getting any clue how to proceed so we use (sec⁡2θ−tan⁡2θ=1)\left( {{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right) to proceed further
=tan⁡θ+sec⁡θ−(sec⁡2θ−tan⁡2θ)tan⁡θ−sec⁡θ+1= \dfrac{{\tan \theta + \sec \theta - \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}{{\tan \theta - \sec \theta + 1}}
Now use property (a2−b2=(a−b)(a+b))\left( {{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right)
= \dfrac{{\tan \theta + \sec \theta - \left\\{ {\left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right)} \right\\}}}{{\tan \theta - \sec \theta + 1}}
Now taking (sec⁡θ+tan⁡θ)\left( {\sec \theta + \tan \theta } \right) common we get
=(tan⁡θ+sec⁡θ)(1−(sec⁡θ−tan⁡θ))tan⁡θ−sec⁡θ+1= \dfrac{{\left( {\tan \theta + \sec \theta } \right)\left( {1 - \left( {\sec \theta - \tan \theta } \right)} \right)}}{{\tan \theta - \sec \theta + 1}}
=(tan⁡θ+sec⁡θ)(1−sec⁡θ+tan⁡θ)(1−sec⁡θ+tan⁡θ)= \dfrac{{\left( {\tan \theta + \sec \theta } \right)\left( {1 - \sec \theta + \tan \theta } \right)}}{{\left( {1 - \sec \theta + \tan \theta } \right)}}
On cancel out we get,
=tan⁡θ+sec⁡θ= \tan \theta + \sec \theta
Now we can write (tan⁡θ=sin⁡θcos⁡θ,sec⁡θ=1cos⁡θ)\left( {\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }},\sec \theta = \dfrac{1}{{\cos \theta }}} \right)
=sin⁡θcos⁡θ+1cos⁡θ =sin⁡θ+1cos⁡θ  = \dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }} \\\ = \dfrac{{\sin \theta + 1}}{{\cos \theta }} \\\
=RHS
Hence Proved.

Note: Whenever we get this type of question the key concept of solving is either we have to start from RHS or LHS and proceed to get required results using standard results. To prove this type of question we have to use the presence of mind that what transformation should tend to require a result.