Question

For $({\theta }_1,{\theta }_2,{\theta }_3,......{\theta }_n)ϵ(0,\pi /2)$ if $ln(sec{\theta }_1-tan{\theta }_1)+ln(sec{\theta }_2-tan{\theta }_2)+....ln(sec\theta -tan{\theta }_n)+ln\pi =0$ then find the value of $cos\left[\right(sec{\theta }_1+tan{\theta }_1\left)\right(sec{\theta }_2+tan{\theta }_2).....(sec{\theta }_n+tan{\theta }_n\left)\right]$

Answer 1

Given $({\theta }_1,{\theta }_2,{\theta }_3....{\theta }_n)\in (\theta ,\frac{\pi }{2})$

$\Rightarrow \ln (\sec {\theta }_1-\tan {\theta }_1)+\ln (\sec {\theta }_2-\tan {\theta }_2)+....+\ln (\sec {\theta }_n-\tan {\theta }_n)+\ln \pi =o....\left(1\right)$

using property of $\log$ we can write

$\Rightarrow \ln \left[\right(\sec {\theta }_1-\tan {\theta }_1\left)\right(\sec {\theta }_2-\tan {\theta }_2)....(\sec {\theta }_n-\tan {\theta }_n\left)\right]=-\ln \pi$

$\Rightarrow$ on dividing each term by it conjucate and multiply we get

$\Rightarrow \ln \left[\frac{({\sec }^2{\theta }_1-{\tan }^2{\theta }_1)({\sec }^2{\theta }_2-{\tan }^2{\theta }_2).....({\sec }^2{\theta }_n-{\tan }^2{\theta }_n)}{(\sec {\theta }_1+\tan {\theta }_1)(\sec {\theta }_2+\tan {\theta }_2)......=-\ln \pi (\sec {\theta }_n+\tan {\theta }_n)}\right]$

$\Rightarrow [\because {\sec }^2\theta -{\tan }^2\theta =1]$

$\Rightarrow \ln \left[\frac{1}{(\sec {\theta }_1+\tan {\theta }_1)(\sec {\theta }_2-\tan {\theta }_2)....(\sec {\theta }_n-\tan {\theta }_n)}\right]=-\ln \pi$

on comparing $LHS$ & $RHS$ we get

$\Rightarrow (\sec {\theta }_1+\tan {\theta }_1)(\sec {\theta }_2+\tan {\theta }_2)....(\sec {\theta }_n+\tan {\theta }_n)=\pi \left(A\right)$

noe we have to find

$\cos \left[\right(\sec {\theta }_1+\tan {\theta }_1\left)\right(\sec {\theta }_2+\tan {\theta }_2)....(\sec {\theta }_n+\tan {\theta }_n\left)\right]$

from equation $\left(1\right)$

$\cos \left[\right(\sec {\theta }_1+\tan {\theta }_1).......(\sec {\theta }_2+\tan {\theta }_2\left)\right]=[\cos \pi =-1]$