If x=log,((cos ((y)/(2))-sin ((y)/(2)))/(cos ((y)/(2))+sin ((y)/(2))))*tan ((y)/(2))=sqrt((1-1)/(1+1)). Then (vi) ,o, has the value (A) (1)/(2) (B) -(1)/(2) (C) (1)/(4) (D) -(1)/(4)

Simplify Right Side: First, let's simplify the right side of the equation. sqrt((1-1)/(1+1)) = sqrt(0/2) = sqrt(0) = 0Analyze Left Side: Now, let's look at the left side of the equation. x = log((cos(y/2) - sin(y/2)) / (cos(y/2) + sin(y/2))) * tan(y/2) Since x = 0, we have: 0 = log((cos(y/2) - sin(y/2)) / (cos(y/2) + sin(y/2))) * tan(y/2)Solve for y - Tan Term: To find y, we need to solve for when the right side equals zero. log((cos(y/2) - sin(y/2)) / (cos(y/2) + sin(y/2))) * tan(y/2) = 0 This can happen if either the log term or the tan term is zero.Solve for y - Log Term: Let's check when tan(y/2) = 0. tan(y/2) = 0 when y/2 = nπ, where n is an integer. But since tan is periodic with π, we only need to consider y/2 = 0 for the smallest solution. So, y = 0 * 2 = 0Correcting Mistake: Now let's check the log term. log((cos(y/2) - sin(y/2)) / (cos(y/2) + sin(y/2))) = 0 This happens when (cos(y/2) - sin(y/2)) / (cos(y/2) + sin(y/2)) = 1Solving the equation (cos(y/2) - sin(y/2)) / (cos(y/2) + sin(y/2)) = 1 cos(y/2) - sin(y/2) = cos(y/2) + sin(y/2) This simplifies to -sin(y/2) = sin(y/2) Which implies sin(y/2) = 0sin(y/2) = 0 when y/2 = mπ, where m is an integer. Again, considering the smallest solution, y/2 = 0. So, y = 0 * 2 = 0Both the tan term and the log term give us y = 0 as a solution. However, we made a mistake in the simplification of the log term; it should not simplify to 1 but to a value that makes the argument of the log equal to 1. We need to correct this.

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If $x=\log\left(\frac{\cos\left(\frac{y}{2}\right)-\sin\left(\frac{y}{2}\right)}{\cos\left(\frac{y}{2}\right)+\sin\left(\frac{y}{2}\right)}\right)\tan\left(\frac{y}{2}\right)=\sqrt{\frac{1-1}{1+1}}$ . Then $(vi)$ , $o$ , has the value $\newline$ (A) $\frac{1}{2}$ $\newline$ (B) $-\frac{1}{2}$ $\newline$ (C) $\frac{1}{4}$ $\newline$ (D) $-\frac{1}{4}$

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Math Problems

Calculus

Find derivatives of logarithmic functions

Full solution

Q. If

x=\log\left(\frac{\cos\left(\frac{y}{2}\right)-\sin\left(\frac{y}{2}\right)}{\cos\left(\frac{y}{2}\right)+\sin\left(\frac{y}{2}\right)}\right)\tan\left(\frac{y}{2}\right)=\sqrt{\frac{1-1}{1+1}}

. Then

(vi)

o

, has the value

\newline

(A)

\frac{1}{2}

\newline

(B)

-\frac{1}{2}

\newline

(C)

\frac{1}{4}

\newline

(D)

-\frac{1}{4}

Simplify Right Side: First, let's simplify the right side of the equation. $\sqrt{\left(\frac{1-1}{1+1}\right)} = \sqrt{\frac{0}{2}} = \sqrt{0} = 0$
Analyze Left Side: Now, let's look at the left side of the equation. $\newline$ $x = \log\left(\frac{\cos(\frac{y}{2}) - \sin(\frac{y}{2})}{\cos(\frac{y}{2}) + \sin(\frac{y}{2})}\right) \cdot \tan(\frac{y}{2})$ $\newline$ Since $x = 0$ , we have: $\newline$ $0 = \log\left(\frac{\cos(\frac{y}{2}) - \sin(\frac{y}{2})}{\cos(\frac{y}{2}) + \sin(\frac{y}{2})}\right) \cdot \tan(\frac{y}{2})$
Solve for $y$ - Tan Term: To find $y$ , we need to solve for when the right side equals zero. $\newline$ $\log\left(\frac{\cos(\frac{y}{2}) - \sin(\frac{y}{2})}{\cos(\frac{y}{2}) + \sin(\frac{y}{2})}\right) \cdot \tan\left(\frac{y}{2}\right) = 0$ $\newline$ This can happen if either the log term or the tan term is zero.
Solve for $y$ - Log Term: Let's check when $\tan(\frac{y}{2}) = 0$ . $\newline$ $\tan(\frac{y}{2}) = 0$ when $\frac{y}{2} = n\pi$ , where $n$ is an integer. $\newline$ But since $\tan$ is periodic with $\pi$ , we only need to consider $\frac{y}{2} = 0$ for the smallest solution. $\newline$ So, $y = 0 \times 2 = 0$
Correcting Mistake: Now let's check the log term. $\log\left(\frac{\cos(\frac{y}{2}) - \sin(\frac{y}{2})}{\cos(\frac{y}{2}) + \sin(\frac{y}{2})}\right) = 0$ This happens when $\frac{\cos(\frac{y}{2}) - \sin(\frac{y}{2})}{\cos(\frac{y}{2}) + \sin(\frac{y}{2})} = 1$
Correcting Mistake: Now let's check the log term. $\newline$ $\log\left(\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)}\right) = 0$ $\newline$ This happens when $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ Solving the equation $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ $\newline$ $\cos(y/2) - \sin(y/2) = \cos(y/2) + \sin(y/2)$ $\newline$ This simplifies to $-\sin(y/2) = \sin(y/2)$ $\newline$ Which implies $\sin(y/2) = 0$
Correcting Mistake: Now let's check the log term. $\newline$ $\log\left(\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)}\right) = 0$ $\newline$ This happens when $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ Solving the equation $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ $\newline$ $\cos(y/2) - \sin(y/2) = \cos(y/2) + \sin(y/2)$ $\newline$ This simplifies to $-\sin(y/2) = \sin(y/2)$ $\newline$ Which implies $\sin(y/2) = 0$ $\sin(y/2) = 0$ when $y/2 = m\pi$ , where $m$ is an integer. $\newline$ Again, considering the smallest solution, $y/2 = 0$ . $\newline$ So, $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ $0$
Correcting Mistake: Now let's check the log term. $\log\left(\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)}\right) = 0$ This happens when $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ Solving the equation $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ $\cos(y/2) - \sin(y/2) = \cos(y/2) + \sin(y/2)$ This simplifies to $-\sin(y/2) = \sin(y/2)$ Which implies $\sin(y/2) = 0$ $\sin(y/2) = 0$ when $y/2 = m\pi$ , where $m$ is an integer. Again, considering the smallest solution, $y/2 = 0$ . So, $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ $0$ Both the tan term and the log term give us $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ $1$ as a solution. However, we made a mistake in the simplification of the log term; it should not simplify to $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ $2$ but to a value that makes the argument of the log equal to $\frac{\cos(y/2) - \sin(y/2)}{\cos(y/2) + \sin(y/2)} = 1$ $2$ . We need to correct this.