Question

Cos-1 ⅘ + cos-1 12/13 = cos-1 33/65

Answer 1

Answer:

Step-by-step explanation:

माना  $L.H.S.=\theta$,  तब  

$\theta>0$और  

$\theta=\cos^{-1}(\dfrac{4}{5} )+cos^{-1}(\dfrac{12}{13} )<\dfrac{\pi}{2} +\dfrac{\pi}{2}\\\\\\=>0<\theta<\pi$

तथा

     $\cos\theta=cos(cos^{-1}(\dfrac{4}{5} )+cos^{-1}(\dfrac{12}{13} ))\\\\\\=cos(cos^{-1}(\dfrac{4}{5} ))\cos(\cos^{-1}(\dfrac{12}{13} ))-\sin(cos^-1}(\dfrac{4}{5} ))\sin(cos^{-1}(\dfrac{12}{13} ))\\\\\\=\dfrac{4}{5} *\dfrac{12}{13}-\sqrt{1-(\dfrac{4}{5})^2} \sqrt{1-(\dfrac{12}{13})^2} \\\\\\=\dfrac{48}{65}-\dfrac{3}{5}*\dfrac{5}{13}\\\\\\=\dfrac{48-15}{65}\\\\=\dfrac{33}{65}\\\\\\\cos\theta=\dfrac{33}{65}=>\theta=\cos^{-1}(\dfrac{33}{65})$