Solveeit Logo
Login

Question

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

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: Try to simplify the left-hand side of the equation that we need to prove by using the formula sec⁡2θ−tan⁡2θ=1{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 , and other similar formula

Complete step-by-step answer:
We will now solve the left-hand side of the equation given in the question.
sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}
Now we will take cos⁡θ\cos \theta common from both the numerator and denominator of the expression. On doing so, we get
=sin⁡θcos⁡θ−1+1cos⁡θsin⁡θcos⁡θ+1−1cos⁡θ=\dfrac{\dfrac{\sin \theta }{\cos \theta }-1+\dfrac{1}{\cos \theta }}{\dfrac{\sin \theta }{\cos \theta }+1-\dfrac{1}{\cos \theta }}
We know that sin⁡xcos⁡x=tan⁡x and 1cos⁡x=sec⁡x\dfrac{\sin x}{\cos x}=\tan x\text{ and }\dfrac{1}{\cos x}=\sec x . Therefore, our expression becomes:
=tan⁡θ−1+sec⁡θtan⁡θ+1−sec⁡θ=\dfrac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta }
Now we know sec⁡2x−tan⁡2x=1{{\sec }^{2}}x-{{\tan }^{2}}x=1 .
=tan⁡θ−1+sec⁡θtan⁡θ+sec⁡2θ−tan⁡2θ−sec⁡θ=\dfrac{\tan \theta -1+\sec \theta }{\tan \theta +{{\sec }^{2}}\theta -{{\tan }^{2}}\theta -\sec \theta }
Now, if we use the formula: a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) , we get
=tan⁡θ−1+sec⁡θsec⁡2θ−tan⁡2θ−(sec⁡θ−tan⁡θ)=\dfrac{\tan \theta -1+\sec \theta }{{{\sec }^{2}}\theta -{{\tan }^{2}}\theta -\left( \sec \theta -\tan \theta \right)}
=tan⁡θ−1+sec⁡θ(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)−(sec⁡θ−tan⁡θ)=\dfrac{\tan \theta -1+\sec \theta }{\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)-\left( \sec \theta -\tan \theta \right)}
=tan⁡θ−1+sec⁡θ(sec⁡θ+tan⁡θ−1)(sec⁡θ−tan⁡θ)=\dfrac{\tan \theta -1+\sec \theta }{\left( \sec \theta +\tan \theta -1 \right)\left( \sec \theta -\tan \theta \right)}
=1sec⁡θ−tan⁡θ=\dfrac{1}{\sec \theta -\tan \theta }
As we have shown that the left-hand side of the equation given in the question is equal to the right-hand side of the equation in the question. Hence, we can say that we have proved 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 } .

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, we need to remember the properties related to complementary angles and trigonometric ratios.