Question

The value of (log_(sqrt(2))(cos20^(@))+log_(sqrt(2))(cos40^(@))+log_(sqrt(2))(cos80^(@)))^(2) +3^(log_((tan60^(@)))(sec60^(@))) is equal to 032sqrt(2)​

Answer 1

$\textbf{To find:}$

$\textsf{The value of}$

$\mathsf{\left(\log_{\sqrt2}cos20^\circ+\log_{\sqrt2}cos40^\circ+\log_{\sqrt2}cos80^\circ\right)^2+3^{\log_{tan60^\circ}sec60^\circ}}$

$\textbf{Solution:}$

$\textbf{Formula used:}$

$\boxed{\mathsf{cos(60^\circ-A)\,cosA\,cos(60^\circ+A)=\dfrac{1}{4}cos3A}}$

$\mathsf{Consider,}$

$\mathsf{cos20^\circ\,cos40^\circ\,cos80^\circ}$

$\mathsf{=cos40^\circ\,cos20^\circ\,cos80^\circ}$

$\mathsf{=cos(60^\circ-20^\circ)\,cos20^\circ\,cos(60^\circ+20^\circ)}$

$\mathsf{=\dfrac{1}{4}\,cos3(20^\circ)}$

$\mathsf{=\dfrac{1}{4}\,cos60^{\circ}}$

$\mathsf{=\dfrac{1}{4}{\times}\dfrac{1}{2}}$

$\mathsf{=\dfrac{1}{8}}$

$\mathsf{Consider,}$

$\mathsf{\left(\log_{\sqrt2}cos20^\circ+\log_{\sqrt2}cos40^\circ+\log_{\sqrt2}cos80^\circ\right)^2+3^{\log_{tan60^\circ}sec60^\circ}}$

$\textsf{Using product rule of logarithm, we get}$

$\mathsf{=\left(\log_{\sqrt2}(cos20^\circ\,cos40^\circ\,cos80^\circ)\right)^2+3^{\log_{\sqrt3}2}}$

$\mathsf{=\left(\log_{\sqrt2}(\frac{1}{8})\right)^2+3^{\log_{\sqrt3}2}}$

$\mathsf{=\left(\log_{\sqrt2}(\frac{1}{\sqrt2})^6\right)^2+3^{\log_{\sqrt3}2}}$

$\mathsf{=\left(\log_{\sqrt2}(\sqrt2)^{-6}\right)^2+(\sqrt{3})^2^{\log_{\sqrt3}2}}$

$\mathsf{=\left(-6\log_{\sqrt2}{\sqrt2}\right)^2+((\sqrt{3})^2)^{\log_{\sqrt3}2}}$

$\mathsf{Using,}\;\boxed{\bf\,\log_aa=1}\;\boxed{\bf\,a^{\log_aN=N}}$

$\mathsf{=\left(-6(1)\right)^2+\sqrt{3}^{2\log_{\sqrt3}2}}$

$\mathsf{=\left(-6\right)^2+\sqrt{3}^{\log_{\sqrt3}2^2}}$

$\mathsf{=36+\sqrt{3}^{\log_{\sqrt3}4}}$

$\mathsf{=36+4}$

$\mathsf{=40}$

$\implies\boxed{\mathsf{\left(\log_{\sqrt2}cos20^\circ+\log_{\sqrt2}cos40^\circ+\log_{\sqrt2}cos80^\circ\right)^2+3^{\log_{tan60^\circ}sec60^\circ}=40}}$

$\textbf{Find more:}$

Log (6x^2-5x+1) base 1-2x - log (4x^2 -4x+1) base 1-3x = 2​

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