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Question: Prove that \[\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\...

Prove that sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=1sec⁡θ−tan⁡θ\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}

Explanation

Solution

Hint: Here R. H. S indicates a term which includes sec⁡θ\sec \theta so, we divide the numerator and denominator of L. H. S by cos⁡θ\cos \theta .

Dividing numerator and denominator by cos⁡θ\cos \theta makes the L. H. S term easy. As sec⁡θ\sec \theta is reciprocal of cos⁡θ\cos \theta so we are dividing L .H .S by cos⁡θ\cos \theta .
⇒\RightarrowL. H. S = sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1÷cos⁡θcos⁡θ\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} \div \dfrac{{\cos \theta }}{{\cos \theta }}

⇒\RightarrowL. H. S =sin⁡θ−cos⁡θ+1cos⁡θsin⁡θ+cos⁡θ−1cos⁡θ\dfrac{{\dfrac{{\sin \theta - \cos \theta + 1}}{{\cos \theta }}}}{{\dfrac{{\sin \theta + \cos \theta - 1}}{{\cos \theta }}}} ……. (1)
Now, by using trigonometric identities we will simplify the L. H. S term.
We know that
sin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta , cos⁡θcos⁡θ=1\dfrac{{\cos \theta }}{{\cos \theta }} = 1, 1cos⁡θ=sec⁡θ\dfrac{1}{{\cos \theta }} = \sec \theta
Now, putting these values in equation (1), we get
⇒\RightarrowL. H. S = tan⁡θ−1+sec⁡θtan⁡θ+1−sec⁡θ\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }} ……. (2)
Now, using the identity sec⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1, equation (2) becomes simple. We replace the value of 1 only in the denominator to get the desired result.
Replacing value of 1 in denominator, we get
⇒\RightarrowL. H. S =tan⁡θ−1+sec⁡θtan⁡θ−sec⁡θ+(sec⁡2θ−tan⁡2θ)\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta - \sec \theta + ({{\sec }^2}\theta - {{\tan }^2}\theta )}}
We know that (a2−b2)=(a−b)(a+b)({a^2} - {b^2}) = (a - b)(a + b), using this property L. H. S can be written as,
⇒\RightarrowL. H. S =tan⁡θ−1+sec⁡θtan⁡θ−sec⁡θ+(sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta - \sec \theta + (\sec \theta - \tan \theta )(\sec \theta + \tan \theta )}}
Taking (sec⁡θ−tan⁡θ)(\sec \theta - \tan \theta ) common from denominator,
⇒\RightarrowL. H. S =tan⁡θ−1+sec⁡θ(sec⁡θ−tan⁡θ)(−1+sec⁡θ+tan⁡θ)\dfrac{{\tan \theta - 1 + \sec \theta }}{{(\sec \theta - \tan \theta )( - 1 + \sec \theta + \tan \theta )}}
Now, eliminating the same terms from the numerator and the denominator, we get
⇒\RightarrowL. H. S =1sec⁡θ−tan⁡θ\dfrac{1}{{\sec \theta - \tan \theta }}= R. H. S
Hence proved.

Note: To solve such problems one should have the knowledge of trigonometric identities like sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 , sec2θ−tan⁡2θ=1se{c^2}\theta - {\tan ^2}\theta = 1 and cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta - {\cot ^2}\theta = 1 which are very useful in solving the problems. Another key point is that we have to see the R. H. S of the question to decide which identity we can use to solve the problem.