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If cosec theta is equal to root 10 find other trigonometric ratios |
| Answer» We have,\xa0{tex}cosec \\;A = \\frac { \\text { Hypotenuse } } { \\text { Perpendicular } } = \\frac { \\sqrt { 10 } } { 1 }{/tex}So, we draw a right triangle ABC, right-angled at B such thatPerpendicular = BC = 1 unit and, Hypotenuse{tex}= A C = \\sqrt { 10 }{/tex}.By Pythagoras theorem, we have{tex}A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }{/tex}{tex}\\Rightarrow \\quad ( \\sqrt { 10 } ) ^ { 2 } = A B ^ { 2 } + 1 ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B ^ { 2 } = 10 - 1 = 9{/tex}{tex}\\Rightarrow \\quad A B = \\sqrt { 9 } = 3{/tex}When we consider the trigonometric ratios of {tex}\\angle A{/tex}, we haveBase = AB = 3 units, Perpendicular = BC = 1 units and, Hypotenuse{tex}= A C = \\sqrt { 10 }{/tex}\xa0units{tex}\\therefore \\quad \\sin A = \\frac { \\text { Perpendicular } } { \\text { Hypotenuse } } = \\frac { 1 } { \\sqrt { 10 } } , \\cos A = \\frac { \\text { Base } } { \\text { Hypotenuse } } = \\frac { 3 } { \\sqrt { 10 } }{/tex}\xa0{tex}\\tan A = \\frac { \\text { Perpendicular } } { \\text { Base } } = \\frac { 1 } { 3 } , \\quad \\sec A = \\frac { \\text { Hypotenuse } } { \\text { Base } } = \\frac { \\sqrt { 10 } } { 3 }{/tex}.and,\xa0{tex}\\cot A = \\frac { \\text { Base } } { \\text { Perpendicular } } = \\frac { 3 } { 1 } = 3{/tex} | |