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

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

Explanation

Solution

The above given question is of trigonometric identities. So, we will use trigonometric identities sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1, and then we will put cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x in the Left-Hand Side and we will use the formula (a2−b2)=(a−b)(a+b)\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) and write (1−sin⁡2x)=(1−sin⁡x)(1+sin⁡x)\left( 1-{{\sin }^{2}}x \right)=\left( 1-\sin x \right)\left( 1+\sin x \right) and then we will simplify it to get the Right-Hand Side of the equation.

Complete step by step answer:
We can see that above given question is of trigonometric identity and so we will use trigonometric identity to prove the above result 1−[cos⁡2x1+sin⁡x]=sin⁡x1-\left[ \dfrac{{{\cos }^{2}}x}{1+\sin x} \right]=\sin x.
Since, we have to prove that 1−[cos⁡2x1+sin⁡x]=sin⁡x1-\left[ \dfrac{{{\cos }^{2}}x}{1+\sin x} \right]=\sin x. So, we will simplify LHS of the equation and make it to RHS i.e. LHS = RHS.
Since, we know that the Left-Hand Side of the equation is 1−[cos⁡2x1+sin⁡x]1-\left[ \dfrac{{{\cos }^{2}}x}{1+\sin x} \right].
And, from trigonometric identity we know that:
⇒sin⁡2x+cos⁡2x=1\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1
Now, we will take cos⁡2x{{\cos }^{2}}x to the right of the equation, then we will get:
⇒cos⁡2x=1−sin⁡2x\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x
So, we will put value of cos⁡2x{{\cos }^{2}}x in LHS term:
⇒1−[1−sin⁡2x1+sin⁡x] ⇒1−[12−sin⁡2x1+sin⁡x] \begin{aligned} & \Rightarrow 1-\left[ \dfrac{1-{{\sin }^{2}}x}{1+\sin x} \right] \\\ & \Rightarrow 1-\left[ \dfrac{{{1}^{2}}-{{\sin }^{2}}x}{1+\sin x} \right] \\\ \end{aligned}
Now, we know that (a2−b2)=(a−b)(a+b)\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right).
So, we can write (1−sin⁡2x)\left( 1-{{\sin }^{2}}x \right) as:
⇒(1−sin⁡2x)=(1−sin⁡x)(1+sin⁡x)\Rightarrow \left( 1-{{\sin }^{2}}x \right)=\left( 1-\sin x \right)\left( 1+\sin x \right)
Now, after putting the value of (1−sin⁡2x)\left( 1-{{\sin }^{2}}x \right) in the above equation we will get:
⇒1−[12−sin⁡2x1+sin⁡x]=1−[(1−sin⁡x)(1+sin⁡x)(1+sin⁡x)]\Rightarrow 1-\left[ \dfrac{{{1}^{2}}-{{\sin }^{2}}x}{1+\sin x} \right]=1-\left[ \dfrac{\left( 1-\sin x \right)\left( 1+\sin x \right)}{\left( 1+\sin x \right)} \right]
Now, we will cancel the common terms of the numerator and the denominator, that is (1+sin⁡x)\left( 1+\sin x \right), then we will get:
⇒1−(1−sin⁡x)\Rightarrow 1-\left( 1-\sin x \right)
Now, we will open the bracket and take minus(-) inside the bracket.
⇒1−1+sin⁡x\Rightarrow 1-1+\sin x
⇒sin⁡x=RHS\Rightarrow \sin x=RHS
Thus, LHS = RHS
Hence, proved.

Note: Students are required to note that when we have to prove any trigonometric equation given in question then we usually make LHS equal to RHS by simplifying anyone side of the given equation to obtain the other side of the equation by using general trigonometric identities.