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Question: How do you prove \(1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=\sin x\)?...

How do you prove 1−cos⁡2x1+sin⁡x=sin⁡x1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=\sin x?

Explanation

Solution

We have sum of two terms in the left-hand side of 1−cos⁡2x1+sin⁡x=sin⁡x1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=\sin x. We use the identities sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 for the numerator. Then we use the identity of a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) to factor the numerator. We complete the division and the binary operations to get the final answer.

Complete step by step solution:
We now use the identity theorem of trigonometry sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 which gives us cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x.
We place the value in the numerator of the equation cos⁡2x1+sin⁡x\dfrac{{{\cos }^{2}}x}{1+\sin x} and get cos⁡2x1+sin⁡x=1−sin⁡2x1+sin⁡x\dfrac{{{\cos }^{2}}x}{1+\sin x}=\dfrac{1-{{\sin }^{2}}x}{1+\sin x}.
The left side of the equation 1−cos⁡2x1+sin⁡x1-\dfrac{{{\cos }^{2}}x}{1+\sin x} gives 1−cos⁡2x1+sin⁡x=1−1−sin⁡2x1+sin⁡x1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=1-\dfrac{1-{{\sin }^{2}}x}{1+\sin x}.
The numerator is in the form of a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right). We now apply the identity theorem for the term 1−sin⁡2x1-{{\sin }^{2}}x. We assume the values a=1,b=sin⁡xa=1,b=\sin x.
Applying the theorem, we get 1−sin⁡2x=(1+sin⁡x)(1−sin⁡x)1-{{\sin }^{2}}x=\left( 1+\sin x \right)\left( 1-\sin x \right).
The equation becomes 1−sin⁡2x1+sin⁡x=(1+sin⁡x)(1−sin⁡x)(1+sin⁡x)\dfrac{1-{{\sin }^{2}}x}{1+\sin x}=\dfrac{\left( 1+\sin x \right)\left( 1-\sin x \right)}{\left( 1+\sin x \right)}.
We can now eliminate the (1+sin⁡x)\left( 1+\sin x \right) from both denominator and numerator.
The equation becomes 1−sin⁡2x1+sin⁡x=(1−sin⁡x)\dfrac{1-{{\sin }^{2}}x}{1+\sin x}=\left( 1-\sin x \right).
The final solution is 1−cos⁡2x1+sin⁡x=1−(1−sin⁡x)=1−1+sin⁡x=sin⁡x1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=1-\left( 1-\sin x \right)=1-1+\sin x=\sin x.
Thus verified 1−cos⁡2x1+sin⁡x=sin⁡x1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=\sin x.

Note: It is important to remember that the condition to eliminate the (1+sin⁡x)\left( 1+\sin x \right) from both denominator and numerator is (1+sin⁡x)≠0\left( 1+\sin x \right)\ne 0. The simplified form is sin⁡x≠−1\sin x\ne -1. No domain is given for the variable xx. The value of sin⁡x≠−1\sin x\ne -1 is essential. The simplified condition will be x≠nπ−(−1)nπ2,n∈Zx\ne n\pi -{{\left( -1 \right)}^{n}}\dfrac{\pi }{2},n\in \mathbb{Z}.