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Question: How do you prove \(\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=2{{\sec }^{...

How do you prove [11−sin⁡x]+[11+sin⁡x]=2sec⁡2x\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=2{{\sec }^{2}}x?

Explanation

Solution

We first simplify the numerator for the addition part of [11−sin⁡x]+[11+sin⁡x]\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right] separately. We have to multiply the denominators for the addition to use it as LCM. We use the identities sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. We also know that the terms cos⁡x\cos x and sec⁡x\sec x are inverse of each other. This gives 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x.

Complete step by step answer:
We have been given an addition of trigonometrical fraction of [11−sin⁡x]+[11+sin⁡x]\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right].
To add them we need to find the LCM of (1−sin⁡x)\left( 1-\sin x \right) and (1+sin⁡x)\left( 1+\sin x \right) which is their multiplication.
So, [11−sin⁡x]+[11+sin⁡x]=(1+sin⁡x)+(1−sin⁡x)(1−sin⁡x)(1+sin⁡x)\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=\dfrac{\left( 1+\sin x \right)+\left( 1-\sin x \right)}{\left( 1-\sin x \right)\left( 1+\sin x \right)}.
We know that (a+b)(a−b)=(a2−b2)\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right). This is a factorisation identity of square subtraction.
Therefore, the LCM is (1−sin⁡x)(1+sin⁡x)=(1−sin⁡2x)\left( 1-\sin x \right)\left( 1+\sin x \right)=\left( 1-{{\sin }^{2}}x \right).
The addition on the numerator is (1+sin⁡x)+(1−sin⁡x)=1+sin⁡x+1−sin⁡x=2\left( 1+\sin x \right)+\left( 1-\sin x \right)=1+\sin x+1-\sin x=2.
So, [11−sin⁡x]+[11+sin⁡x]=2(1−sin⁡2x)\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=\dfrac{2}{\left( 1-{{\sin }^{2}}x \right)}.
For the numerator part we have 1−sin⁡2x=cos⁡2x1-{{\sin }^{2}}x={{\cos }^{2}}x as for any value of xx, sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1.
Therefore, [11−sin⁡x]+[11+sin⁡x]=2cos⁡2x\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=\dfrac{2}{{{\cos }^{2}}x}
We know that the terms cos⁡x\cos x and sec⁡x\sec x are inverse of each other. So, 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x.
Squaring we get 1cos⁡2x=sec⁡2x\dfrac{1}{{{\cos }^{2}}x}={{\sec }^{2}}x. We put the value in the equation to get
[11−sin⁡x]+[11+sin⁡x]=2×1cos⁡2x=2sec⁡2x\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right]=2\times \dfrac{1}{{{\cos }^{2}}x}=2{{\sec }^{2}}x.
Therefore, the simplified solution of [11−sin⁡x]+[11+sin⁡x]\left[ \dfrac{1}{1-\sin x} \right]+\left[ \dfrac{1}{1+\sin x} \right] is 2sec⁡2x2{{\sec }^{2}}x. Thus, verified.

Note: We need to remember that the final terms are square terms. The identities sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 and 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x are valid for any value of xx. In the case of 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x, the only condition is cos⁡x≠0\cos x\ne 0. The division of the fraction part only gives 1 as the solution.