Search any question & find its solution
Question:
Answered & Verified by Expert
In a right-angled triangle, the sides are $a, b$ and $c$, with $c$ as hypotenuse, and $c-b \neq 1, c+b \neq 1$. Then the value of $\left(\log _{c+b} a+\log _{c-b} a\right) /\left(2 \log _{c+b} a \times \log _{c-b} a\right)$ will be
Options:
Solution:
1601 Upvotes
Verified Answer
The correct answer is:
1
Hints : $c^2=a^2+b^2$
$\Rightarrow \mathrm{c}^2-\mathrm{b}^2=\mathrm{a}^2$
$\frac{\frac{\log a}{\log (c+b)}+\frac{\log a}{\log (c-b)}}{\frac{2 \log a \times \log a}{\log (c+b) \log (c-b)}}=\frac{\log a\left(\log \left(c^2-b^2\right)\right)}{2 \log a \log a}=\frac{\log a^2}{\log a^2}=1$
$\Rightarrow \mathrm{c}^2-\mathrm{b}^2=\mathrm{a}^2$
$\frac{\frac{\log a}{\log (c+b)}+\frac{\log a}{\log (c-b)}}{\frac{2 \log a \times \log a}{\log (c+b) \log (c-b)}}=\frac{\log a\left(\log \left(c^2-b^2\right)\right)}{2 \log a \log a}=\frac{\log a^2}{\log a^2}=1$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.