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Question: Simplify: \( \dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}} \)...

Simplify: tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1\dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}}

Explanation

Solution

Hint : Use different identities of trigonometric properties. Also, use the method of factorization for quadratic expansions. Use formulas like (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2} and tangent and secant relation equation. sec⁡2θ=1+tan⁡2θ{\sec ^2}\theta = 1 + {\tan ^2}\theta

Complete step-by-step answer :
Take the given expression –
tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1\dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}}
The above expression can be re-written as –
tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1=tan⁡θ−1+sec⁡θtan⁡θ+1−sec⁡θ\dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}} = \dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}
Again, we can write terms into brackets –
=(tan⁡θ−1)+sec⁡θ(tan⁡θ+1)−sec⁡θ= \dfrac{{(\tan \theta - 1) + \sec \theta }}{{(\tan \theta + 1) - \sec \theta }}
Now, using the property that multiplying and dividing the fraction with the same terms gives the resultant value as the equivalent value of that fraction.
So, here we are multiplying and dividing the above expression by (tan⁡θ+1)+sec⁡θ(\tan \theta + 1) + \sec \theta
=(tan⁡θ−1)+sec⁡θ(tan⁡θ+1)−sec⁡θ×(tan⁡θ+1)+sec⁡θ(tan⁡θ+1)+sec⁡θ= \dfrac{{(\tan \theta - 1) + \sec \theta }}{{(\tan \theta + 1) - \sec \theta }} \times \dfrac{{(\tan \theta + 1) + \sec \theta }}{{(\tan \theta + 1) + \sec \theta }}
Multiply and simplify the above expression. Also, in the above expression we can observe that the denominator terms can be expressed as the difference of squares formula. i.e. (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}
=(tan⁡θ−1)(tan⁡θ+1)+sec⁡θ(tan⁡θ+1)+(tan⁡θ−1)sec⁡θ+sec⁡θ(sec⁡θ)(tan⁡θ+1)(tan⁡θ+1)−(sec⁡θ)(sec⁡θ)= \dfrac{{(\tan \theta - 1)(\tan \theta + 1) + \sec \theta (\tan \theta + 1) + (\tan \theta - 1)\sec \theta + \sec \theta (\sec \theta )}}{{(\tan \theta + 1)(\tan \theta + 1) - (\sec \theta )(\sec \theta )}}
Expand the above expression by opening the brackets.
=(tan⁡2θ−12)+sec⁡θtan⁡θ+sec⁡θ‾+tan⁡θsec⁡θ−sec⁡θ‾+sec⁡2θ(tan⁡θ+1)2−(sec⁡2θ)= \dfrac{{({{\tan }^2}\theta - {1^2}) + \sec \theta \tan \theta \overline { + \sec \theta } + \tan \theta \sec \theta \overline { - \sec \theta } + {{\sec }^2}\theta }}{{{{(\tan \theta + 1)}^2} - ({{\sec }^2}\theta )}}
Terms with the same value and opposite sign cancel each other in the numerator of the above expression.
=tan⁡2θ−1+sec⁡θtan⁡θ+tan⁡θsec⁡θ‾+sec⁡2θ(tan⁡θ+1)2−(sec⁡2θ)= \dfrac{{{{\tan }^2}\theta - 1 + \underline {\sec \theta \tan \theta + \tan \theta \sec \theta } + {{\sec }^2}\theta }}{{{{(\tan \theta + 1)}^2} - ({{\sec }^2}\theta )}}
Add and subtract among the like terms in the above expressions.
=tan⁡2θ−1+2tan⁡θsec⁡θ+sec⁡2θtan⁡2θ+2tan⁡θ+1−sec⁡2θ= \dfrac{{{{\tan }^2}\theta - 1 + 2\tan \theta \sec \theta + {{\sec }^2}\theta }}{{{{\tan }^2}\theta + 2\tan \theta + 1 - {{\sec }^2}\theta }}
Place sec⁡2θ=1+tan⁡2θ{\sec ^2}\theta = 1 + {\tan ^2}\theta in the above expression –
=tan⁡2θ−1+2tan⁡θsec⁡θ+1+tan⁡2θtan⁡2θ+2tan⁡θ+1−(1+tan⁡2θ)= \dfrac{{{{\tan }^2}\theta - 1 + 2\tan \theta \sec \theta + 1 + {{\tan }^2}\theta }}{{{{\tan }^2}\theta + 2\tan \theta + 1 - (1 + {{\tan }^2}\theta )}}
Remember when you open the brackets and sign outside the bracket is a negative sign of all the terms inside the bracket also changing. Negative turns to positive and positive turns to negative.
=tan⁡2θ−1+2tan⁡θsec⁡θ+1+tan⁡2θtan⁡2θ+2tan⁡θ+1−1−tan⁡2θ= \dfrac{{{{\tan }^2}\theta - 1 + 2\tan \theta \sec \theta + 1 + {{\tan }^2}\theta }}{{{{\tan }^2}\theta + 2\tan \theta + 1 - 1 - {{\tan }^2}\theta }}
Terms with the same value and opposite sign cancel each other.
=tan⁡2θ+2tan⁡θsec⁡θ+tan⁡2θ2tan⁡θ= \dfrac{{{{\tan }^2}\theta + 2\tan \theta \sec \theta + {{\tan }^2}\theta }}{{2\tan \theta }}
Add the like terms
=2tan⁡2θ+2tan⁡θsec⁡θ2tan⁡θ= \dfrac{{2{{\tan }^2}\theta + 2\tan \theta \sec \theta }}{{2\tan \theta }}
Take common multiple from the numerator.
=2tan⁡θ(tan⁡θ+sec⁡θ)2tan⁡θ= \dfrac{{2\tan \theta (\tan \theta + \sec \theta )}}{{2\tan \theta }}
Common factors from the numerator and the denominator cancel each other.
=(tan⁡θ+sec⁡θ)= (\tan \theta + \sec \theta )
Hence the required simplified form –
tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1=(tan⁡θ+sec⁡θ)\dfrac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}} = (\tan \theta + \sec \theta )
So, the correct answer is “(tan⁡θ+sec⁡θ)(\tan \theta + \sec \theta ) ”.

Note : The above expression can be solved by substituting all the terms and converting them into sine and cosine functions. Also, remember that the most important property of sines and cosines is that their values lie between minus one and plus one. Every point on the circle is unit circle from the origin. So, the coordinates of any point are within one of zero as well.
Directly the Pythagoras identity are followed by sines and cosines which concludes that –
Sin⁡2θ+Cos⁡2θ=1{\operatorname{Sin} ^2}\theta + {\operatorname{Cos} ^2}\theta = 1