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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=1secθ−tan⁡θ\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{sec\theta - \tan \theta }} using sec2θ=1+tan⁡2θse{c^2}\theta = 1 + {\tan ^2}\theta

Explanation

Solution

Hint : On the left hand side of the given equation, we have sine and cosine; on the right hand side, we have tangent and secant. So divide the numerator and denominator of the left hand side of the equation by cos⁡θ\cos \theta . We will get a result in terms of tangent and secant as tangent is the ratio of sine and cosine; secant is the inverse of cosine. Using sec2θ=1+tan⁡2θse{c^2}\theta = 1 + {\tan ^2}\theta , prove the given equation.

Complete step-by-step answer :
We are given to prove sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=1secθ−tan⁡θ\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{sec\theta - \tan \theta }} .
And sec2θ=1+tan⁡2θse{c^2}\theta = 1 + {\tan ^2}\theta
Sending tan⁡2θ{\tan ^2}\theta from the right hand side to left hand side, we get
sec2θ−tan⁡2θ=1se{c^2}\theta - {\tan ^2}\theta = 1 …….. equation (1)
Considering secθsec\theta as ‘a’ and tan⁡θ\tan \theta as ‘b’, the above equation becomes a2−b2=1{a^2} - {b^2} = 1
We already know that a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
Therefore, sec2θ−tan⁡2θ=(secθ+tan⁡θ)(secθ−tan⁡θ)se{c^2}\theta - {\tan ^2}\theta = \left( {sec\theta + \tan \theta } \right)\left( {sec\theta - \tan \theta } \right) …… equation (2)
The given equation to prove is sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=1secθ−tan⁡θ\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{sec\theta - \tan \theta }}
Considering the LHS
⇒sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1\Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}}
We are now dividing the numerator and denominator of the above trigonometric expression by cos⁡θ\cos \theta
⇒(sin⁡θ−cos⁡θ+1cos⁡θ)(sin⁡θ+cos⁡θ−1cos⁡θ)\Rightarrow \dfrac{{\left( {\dfrac{{\sin \theta - \cos \theta + 1}}{{\cos \theta }}} \right)}}{{\left( {\dfrac{{\sin \theta + \cos \theta - 1}}{{\cos \theta }}} \right)}}
⇒(sin⁡θcos⁡θ−cos⁡θcos⁡θ+1cos⁡θ)(sin⁡θcos⁡θ+cos⁡θcos⁡θ−1cos⁡θ)\Rightarrow \dfrac{{\left( {\dfrac{{\sin \theta }}{{\cos \theta }} - \dfrac{{\cos \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }}} \right)}}{{\left( {\dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\cos \theta }} - \dfrac{1}{{\cos \theta }}} \right)}}
We know that the ratio of sine and cosine gives tan and the inverse of cosine is secant.
This gives,
⇒tan⁡θ−1+secθtan⁡θ+1−secθ\Rightarrow \dfrac{{\tan \theta - 1 + sec\theta }}{{\tan \theta + 1 - sec\theta }}
Writing tangent and secant terms together
⇒tan⁡θ+secθ−1tan⁡θ−secθ+1\Rightarrow \dfrac{{\tan \theta + sec\theta - 1}}{{\tan \theta - sec\theta + 1}}
Now we are multiplying the numerator and denominator by tan⁡θ−secθ\tan \theta - sec\theta
⇒(tan⁡θ+secθ−1)(tan⁡θ−secθ)(tan⁡θ−secθ+1)(tan⁡θ−secθ)\Rightarrow \dfrac{{\left( {\tan \theta + sec\theta - 1} \right)\left( {\tan \theta - sec\theta } \right)}}{{\left( {\tan \theta - sec\theta + 1} \right)\left( {\tan \theta - sec\theta } \right)}}
⇒(tan⁡θ+secθ)(tan⁡θ−secθ)−(tan⁡θ−secθ)(tan⁡θ−secθ)(tan⁡θ−secθ)+(tan⁡θ−secθ)\Rightarrow \dfrac{{\left( {\tan \theta + sec\theta } \right)\left( {\tan \theta - sec\theta } \right) - \left( {\tan \theta - sec\theta } \right)}}{{\left( {\tan \theta - sec\theta } \right)\left( {\tan \theta - sec\theta } \right) + \left( {\tan \theta - sec\theta } \right)}}
⇒(tan⁡2θ−sec2θ)−(tan⁡θ−secθ)(tan⁡θ−secθ)2+(tan⁡θ−secθ)\Rightarrow \dfrac{{\left( {{{\tan }^2}\theta - se{c^2}\theta } \right) - \left( {\tan \theta - sec\theta } \right)}}{{{{\left( {\tan \theta - sec\theta } \right)}^2} + \left( {\tan \theta - sec\theta } \right)}} ( since from equation 2)
We know from equation 1 that sec2θ−tan⁡2θ=1se{c^2}\theta - {\tan ^2}\theta = 1 , this means tan⁡2θ−sec2θ=−1{\tan ^2}\theta - se{c^2}\theta = - 1
⇒−1−(tan⁡θ−secθ)(tan⁡θ−secθ)2+(tan⁡θ−secθ)\Rightarrow \dfrac{{ - 1 - \left( {\tan \theta - sec\theta } \right)}}{{{{\left( {\tan \theta - sec\theta } \right)}^2} + \left( {\tan \theta - sec\theta } \right)}}
⇒−[1+(tan⁡θ−secθ)](tan⁡θ−secθ)(1+(tan⁡θ−secθ))\Rightarrow \dfrac{{ - \left[ {1 + \left( {\tan \theta - sec\theta } \right)} \right]}}{{\left( {\tan \theta - sec\theta } \right)\left( {1 + \left( {\tan \theta - sec\theta } \right)} \right)}}
Cancelling 1+(tan⁡θ−secθ)1 + \left( {\tan \theta - sec\theta } \right) in the numerator and denominator, we get
⇒−1(tan⁡θ−secθ)\Rightarrow \dfrac{{ - 1}}{{\left( {\tan \theta - sec\theta } \right)}}
On multiplying the numerator and denominator by -1, we get
⇒1(secθ−tan⁡θ)\Rightarrow \dfrac{1}{{\left( {sec\theta - \tan \theta } \right)}}
Therefore, the value of sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} is 1secθ−tan⁡θ\dfrac{1}{{sec\theta - \tan \theta }} .
Hence, proved.

Note : Here we have considered the LHS of the equation and proved it that it is equal to the RHS. We can also prove it by first considering the RHS and finding its solution. sec2θ−tan⁡2θ=1se{c^2}\theta - {\tan ^2}\theta = 1 is one of the Pythagorean identities. sin⁡2θ+cos⁡2θ=1,cos⁡ec2θ−cot2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1,\cos e{c^2}\theta - co{t^2}\theta = 1 are the other two Pythagorean identities.