Question

Find the value of (sin 780 )(sin 480°) – (cos120°) (sin 150°) + (sec 610 )(cosec160°) – (cot380) (tan 470°)​

Answer 1

$\textbf{To find:}$

$\text{The value of}$

$sin780^{\circ}\;sin480^{\circ}-cos120^{\circ}\;sin150^{\circ}+sec610^{\circ}\;cosec160^{\circ}-cot380^{\circ}\;tan470^{\circ}$

$sin780^{\circ}=sin(720^{\circ}+60^{\circ})=sin60^{\circ}=\frac{\sqrt{3}}{2}$

$sin480^{\circ}=sin(360^{\circ}+120^{\circ})=sin120^{\circ}=cos30^{\circ}=\frac{\sqrt{3}}{2}$

$cos120^{\circ}=-sin30^{\circ}=\frac{-1}{2}$

$sin150^{\circ}=cos60^{\circ}=\frac{1}{2}$

$sec610^{\circ}=sec(360^{\circ}+250^{\circ})=sec250^{\circ}=-sec70^{\circ}$

$cossec160^{\circ}=sec70^{\circ}$

$cot380^{\circ}=cot20^{\circ}$

$tan470^{\circ}=tan(360^{\circ}+110^{\circ})=tan110^{\circ}=-cot20^{\circ}$

$\text{Now,}$

$\text{The value of}$

$sin780^{\circ}\;sin480^{\circ}-cos120^{\circ}\;sin150^{\circ}+sec610^{\circ}\;cosec160^{\circ}-cot380^{\circ}\;tan470^{\circ}$

$=(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})-(\frac{-1}{2})(\frac{1}{2})+(-sec70^{\circ})(sec70^{\circ})-(cot20^{\circ})(-cot20^{\circ})$

$=\dfrac{3}{4}+\dfrac{1}{4}-sec^270^{\circ}+cot^220^{\circ}$

$=\dfrac{3}{4}+\dfrac{\sqrt{1}}{4}-sec^270^{\circ}+tan^270^{\circ}$

$=\dfrac{3}{4}+\dfrac{1}{4}-(sec^270^{\circ}-tan^270^{\circ})$

$=\dfrac{3}{4}+\dfrac{1}{4}-(1)$

$=1-1$

$=0$

$\therefore\textbf{The value of }$

$sin780^{\circ}\;sin480^{\circ}-cos120^{\circ}\;sin150^{\circ}+sec610^{\circ}\;cosec160^{\circ}-cot380^{\circ}\;tan470^{\circ}\;\text{is}\;\bf\,0$

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