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

seo-qna
Last updated date: 18th Jun 2024
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Answer
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Hint: We have sum of two terms in the left-hand side of $1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=\sin x$. We use the identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ for the numerator. Then we use the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ 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 ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ which gives us ${{\cos }^{2}}x=1-{{\sin }^{2}}x$.
We place the value in the numerator of the equation $\dfrac{{{\cos }^{2}}x}{1+\sin x}$ and get $\dfrac{{{\cos }^{2}}x}{1+\sin x}=\dfrac{1-{{\sin }^{2}}x}{1+\sin x}$.
The left side of the equation $1-\dfrac{{{\cos }^{2}}x}{1+\sin x}$ gives $1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=1-\dfrac{1-{{\sin }^{2}}x}{1+\sin x}$.
The numerator is in the form of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We now apply the identity theorem for the term $1-{{\sin }^{2}}x$. We assume the values $a=1,b=\sin x$.
Applying the theorem, we get $1-{{\sin }^{2}}x=\left( 1+\sin x \right)\left( 1-\sin x \right)$.
The equation becomes $\dfrac{1-{{\sin }^{2}}x}{1+\sin x}=\dfrac{\left( 1+\sin x \right)\left( 1-\sin x \right)}{\left( 1+\sin x \right)}$.
We can now eliminate the $\left( 1+\sin x \right)$ from both denominator and numerator.
The equation becomes $\dfrac{1-{{\sin }^{2}}x}{1+\sin x}=\left( 1-\sin x \right)$.
The final solution is $1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=1-\left( 1-\sin x \right)=1-1+\sin x=\sin x$.
Thus verified $1-\dfrac{{{\cos }^{2}}x}{1+\sin x}=\sin x$.

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