Math, asked by vansh7638, 8 months ago

find the value of-
(sin 780°)(sin 480)-(cos 120°) ( sin 150) + (sec 610)(cosec160)-(cot 380)(tan 470)

Answers

Answered by MaheswariS
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{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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