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

Prove that cos⁡θ+sin⁡θ+1sin⁡θ+cos⁡θ−1=cos⁡ecθ+cot⁡θ\dfrac{\cos \theta +\sin \theta +1}{\sin \theta +\cos \theta -1}=\cos ec\theta +\cot \theta

Explanation

Solution

Hint: Try to simplify the left-hand side of the equation that we need to prove by using the formula cosec2θ−cot⁡2θ=1\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1 and other similar formulas.

Complete step-by-step answer:
Before moving to the solution, let us discuss the periodicity of the secant and tangent function which we would be using in the solution. All the trigonometric ratios including secant and tangent are periodic functions. We can better understand this using the graph of secant and tangent.
First, let us start with the graph of cosec x.

Next, let us see the graph of cotx.

Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. 2πc2{{\pi }^{c}} . So, we can say that the fundamental period of the secant function and the tangent function is 2πc=360∘2{{\pi }^{c}}=360{}^\circ
We will now solve the left-hand side of the equation given in the question.
cos⁡θ−sin⁡θ+1sin⁡θ+cos⁡θ−1\dfrac{\cos \theta -\sin \theta +1}{\sin \theta +\cos \theta -1}
Now we will take sin⁡θ\sin \theta common from both the numerator and denominator of the expression. On doing so, we get
cos⁡θsin⁡θ−1+1sin⁡θcos⁡θsin⁡θ+1−1sin⁡θ\dfrac{\dfrac{\cos \theta }{\sin \theta }-1+\dfrac{1}{\sin \theta }}{\dfrac{\cos \theta }{\sin \theta }+1-\dfrac{1}{\sin \theta }}
We know that cos⁡xsinx=cot⁡x and 1sin⁡x=cos⁡ecx\dfrac{\cos x}{sinx}=\cot x\text{ and }\dfrac{1}{\sin x}=\cos ecx . Therefore, our expression becomes:
cot⁡θ−1+cos⁡ecθcot⁡θ+1−cos⁡ecθ\dfrac{\cot \theta -1+\cos ec\theta }{\cot \theta +1-\cos ec\theta }
Now we know cos⁡ec2x−cot⁡2x=1\cos e{{c}^{2}}x-{{\cot }^{2}}x=1 .
cot⁡θ−(cos⁡ec2θ−cot⁡2θ)+cos⁡ecθcot⁡θ+1−cos⁡ecθ\dfrac{\cot \theta -\left( \cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta \right)+\cos ec\theta }{\cot \theta +1-\cos ec\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
cos⁡ecθ+cot⁡θ−(cos⁡ec2θ−cot⁡2θ)cot⁡θ+1−cos⁡ecθ\dfrac{\cos ec\theta +\cot \theta -\left( \cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta \right)}{\cot \theta +1-\cos ec\theta }
=cos⁡ecθ+cot⁡θ−(cos⁡ecθ−cot⁡θ)(cos⁡ecθ+cot⁡θ)cot⁡θ+1−cos⁡ecθ=\dfrac{\cos ec\theta +\cot \theta -\left( \cos ec\theta -\cot \theta \right)\left( \cos ec\theta +\cot \theta \right)}{\cot \theta +1-\cos ec\theta }
=(cos⁡ecθ+cot⁡θ)(1−cos⁡ecθ+cot⁡θ)cot⁡θ+1−cos⁡ecθ=\dfrac{\left( \cos ec\theta +\cot \theta \right)\left( 1-\cos ec\theta +{{\cot }}\theta \right)}{\cot \theta +1-\cos ec\theta }
=cos⁡ecθ+cot⁡θ=\cos ec\theta +\cot \theta
As we have shown that left-hand side of the equation given in the question is equal to right-hand side of the equation in the question. Hence, we can say that we have proved that cos⁡θ+sin⁡θ+1sin⁡θ+cos⁡θ−1=cos⁡ecθ+cot⁡θ\dfrac{\cos \theta +\sin \theta +1}{\sin \theta +\cos \theta -1}=\cos ec\theta +\cot \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. Important step is to replace the value of 1. Also, you need to remember the properties related to complementary angles and trigonometric ratios.