If for x in left(0, frac{pi}{2}right),log_{10 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $x \in \left(0, \frac{\pi}{2}\right)$.From the first equation: $\log_{10} \sin x + \log_{10} \cos x = -1$$\Rightarrow \log_{10}(\sin x \cos

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Given $x \in \left(0, \frac{\pi}{2}ight)$.From the first equation: $\log_{10} \sin x + \log_{10} \cos x = -1$$\Rightarrow \log_{10}(\sin x \cos x) = -1$$\Rightarrow \sin x \cos x = 10^{-1} = \frac{1}{10} \quad ....(1)$From the second equation: $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) = \frac{1}{2} \log_{10} \left(\frac{n}{10}ight) = \log_{10} \sqrt{\frac{n}{10}}$$\Rightarrow \sin x + \cos x = \sqrt{\frac{n}{10}}$Squaring both sides:$(\sin x + \cos x)^2 = \frac{n}{10}$$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{n}{10}$$1 + 2 \left(\frac{1}{10}ight) = \frac{n}{10}$$1 + \frac{1}{5} = \frac{n}{10}$$\frac{6}{5} = \frac{n}{10}$$n = \frac{6 	imes 10}{5} = 12$.

If for $x \in \left(0, \frac{\pi}{2}\right)$,$\log_{10} \sin x + \log_{10} \cos x = -1$ and $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$,$n > 0$,then the value of $n$ is equal to

Similar Questions

The value of $0.\overline{234}$ is

The value of $a^{\log_b x}$,where $a = 0.2$,$b = \sqrt{5}$,and $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ to $\infty$ is:

The number of solution pairs $(x, y)$ of the simultaneous equations $\log _{1 / 3}(x+y)+\log _3(x-y)=2$ and $2^{y^2}=512^{x+1}$ is

If ${a^x} = {(x + y + z)^y}$,${a^y} = {(x + y + z)^z}$,and ${a^z} = {(x + y + z)^x}$,then:

The square root of $\frac{(0.75)^3}{1-0.75}+[0.75+(0.75)^2+1]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module