Question

soution Let ( cos ^{-1} frac{4}{5}=theta ) Then, ( cos theta=frac{4}{5} )
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therefore quad tan theta=frac{sqrt{1-cos ^{2} theta}}{cos theta}=frac{sqrt{1-frac{16}{25}}}{(4 / 5)}=left(frac{3}{5} times frac{5}{4}right)=frac{3}{4}
Rightarrow theta=tan ^{-1} frac{3}{4}
therefore cos ^{-1} frac{4}{5}=tan ^{-1} frac{3}{4}
therefore quad cos ^{-1} frac{4}{5}+tan ^{-1} frac{3}{5}=tan ^{-1} frac{3}{4}+tan ^{-1} frac{3}{5}
=tan ^{-1}left(frac{frac{3}{4}+frac{3}{5}}{1-frac{3}{4} times frac{3}{5}}right)=tan ^{-1} frac{27}{11}
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# EXAMPLE 20 Prove that cos 14 + tan-1 3 = tan - 27 COS 11

Solution