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

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

Explanation

Solution

First we have to define what the terms we need to solve the problem are.
In trigonometric we studied many formulas using the angles and sin, cos, tan, sec, cosec, and cot.
Since sec is the inverse of sin and cosec is the inverse of cos also cot is the inverse of tan. If we divide the sin and cos, we get tan theta. Hence all the formulas are interrelated in the trigonometric functions.
Here in this problem, we had another expression of the trigonometric function and we need to prove both sides are equal.
Formula used: (cos⁡ecθ−cot⁡θ)(cos⁡ecθ+cot⁡θ)=cos⁡ec2θ−cot⁡2θ(\cos ec\theta - \cot \theta )(\cos ec\theta + \cot \theta ) = \cos e{c^2}\theta - {\cot ^2}\theta and 1sin⁡θ=cos⁡ecθ\dfrac{1}{{\sin \theta }} = \cos ec\theta .

Complete step by step answer:
Since there are two expressions from the given question.
Let us take the left-hand side expression first, which is 1cos⁡ecθ−cot⁡θ−1sin⁡θ\dfrac{1}{{\cos ec\theta - \cot \theta }} - \dfrac{1}{{\sin \theta }} .
As we know the method of conjugation which is denominator terms yields positive values.
Hence by the conjugation, we get the expression as; 1cos⁡ecθ−cot⁡θ×cos⁡ecθ+cot⁡θcos⁡ecθ+cot⁡θ−1sin⁡θ\dfrac{1}{{\cos ec\theta - \cot \theta }} \times \dfrac{{\cos ec\theta + \cot \theta }}{{\cos ec\theta + \cot \theta }} - \dfrac{1}{{\sin \theta }} (since if we cancel both values the expression cannot be changed).
Thus apply the formula given in the hint we get; cos⁡ecθ+cot⁡θcos⁡ec2θ−cot⁡2θ−1sin⁡θ\dfrac{{\cos ec\theta + \cot \theta }}{{\cos e{c^2}\theta - {{\cot }^2}\theta }} - \dfrac{1}{{\sin \theta }} and also 1sin⁡θ=cos⁡ecθ\dfrac{1}{{\sin \theta }} = \cos ec\theta .
Therefore, we get cos⁡ecθ+cot⁡θcos⁡ec2θ−cot⁡2θ−cos⁡ecθ\dfrac{{\cos ec\theta + \cot \theta }}{{\cos e{c^2}\theta - {{\cot }^2}\theta }} - \cos ec\theta . since, cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta - {\cot ^2}\theta = 1 as like as the sin and cos square terms. Thus, we get the expression as cos⁡ecθ+cot⁡θ−cos⁡ecθ\cos ec\theta + \cot \theta - \cos ec\theta (denominator term is one).
Further solving this we get;cos⁡ecθ+cot⁡θ−cos⁡ecθ=cot⁡θ\cos ec\theta + \cot \theta - \cos ec\theta = \cot \theta (common terms canceled).
Hence, we get the left-hand side equation as 1cos⁡ecθ−cot⁡θ−1sin⁡θ=cot⁡θ\dfrac{1}{{\cos ec\theta - \cot \theta }} - \dfrac{1}{{\sin \theta }} = \cot \theta .
Similarly, for the right-hand side expression we use the same methods as follows; 1sin⁡θ−1cos⁡ecθ−cot⁡θ\dfrac{1}{{\sin \theta }} - \dfrac{1}{{\cos ec\theta - \cot \theta }} can be rewritten by the conjugation method 1sin⁡θ−1cos⁡ecθ−cot⁡θ×cos⁡ecθ+cot⁡θcos⁡ecθ+cot⁡θ\dfrac{1}{{\sin \theta }} - \dfrac{1}{{\cos ec\theta - \cot \theta }} \times \dfrac{{\cos ec\theta + \cot \theta }}{{\cos ec\theta + \cot \theta }} and applying the formulas in the hint; we get cos⁡ecθ−cos⁡ecθ+cot⁡θcos⁡ec2θ−cot⁡2θ\cos ec\theta - \dfrac{{\cos ec\theta + \cot \theta }}{{\cos e{c^2}\theta - {{\cot }^2}\theta }} and also cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta - {\cot ^2}\theta = 1 .
Thus, we get cos⁡ecθ+cot⁡θ−cos⁡ecθ=cot⁡θ\cos ec\theta + \cot \theta - \cos ec\theta = \cot \theta and hence also in the right-hand side expression we get 1sin⁡θ−1cos⁡ecθ−cot⁡θ=cot⁡θ\dfrac{1}{{\sin \theta }} - \dfrac{1}{{\cos ec\theta - \cot \theta }} = \cot \theta .
Therefore, on both sides of the expression we get cot⁡θ\cot \theta and hence the left-hand side equals the right-hand side.

Note: Since conjugation is the method of converting the terms into positive values in the denominator for trigonometry.
But for the complex numbers, these conjugations will be changed like ii is the conjugation of −i- i .
Also cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta - {\cot ^2}\theta = 1 , as it will replicate the sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 .