If cot theta =3\\4, show that[sectheta -cosec theta \\SEC Theta +cosec theta]in root=1\\root7

If cot theta =3\\4, show that[sectheta -cosec theta \\SEC Theta +cosec

1.	If cot theta =3\\4, show that[sectheta -cosec theta \\SEC Theta +cosec theta]in root=1\\root7
Answer» Let us draw a triangle ABC such that,\xa0{tex}\\angle{/tex}B =\xa090°.Let\xa0{tex}\\angle{/tex}A =\xa0{tex}\\theta{/tex}°.We have,\xa0{tex}\\cot \\theta = \\frac { 3 } { 4 }{/tex}Then,\xa0{tex}\\cot \\theta = \\frac { \\text { Base } } { \\text { Perpendiaular } } = \\frac { A B } { B C } = \\frac { 3 } { 4 }{/tex}Let AB\xa0= 3 and BC = 4,By Pythagoras\' theorem, we know thatAC2\xa0= AB2 + BC2= 32\xa0+ 42\xa0= 9\xa0+ 16\xa0= 25{tex}\\Rightarrow \\quad AC = 5{/tex}Now,{tex}\\sec \\theta = \\frac { \\text { Hypotenuse } } { \\text { Base } } = \\frac { A C } { A B } = \\frac { 5 } { 3 }{/tex}{tex}\\text{cosec} \\theta = \\frac { \\text { Hypotenuse } } { \\text { Perpendicular } } = \\frac { A C } { B C } = \\frac { 5 } { 4 }{/tex}L.H.S =\xa0{tex}\\sqrt { \\frac { \\sec \\theta - \\text{cosec} \\theta } { \\sec \\theta + \\text{cosec} \\theta } }{/tex}{tex}= \\sqrt { \\frac { 5 / 3 - 5 / 4 } { 5 / 3 + 5 / 4 } }{/tex}{tex}= \\sqrt { \\frac { \\frac { 20 - 15 } { 12 } } { \\frac { 20 + 15 } { 12 } } }{/tex}{tex}= \\sqrt { \\frac { 5 } { 35 } }{/tex}{tex}= \\sqrt { \\frac { 1 } { 7 } }{/tex}{tex}= \\frac { 1 } { \\sqrt { 7 } }{/tex}= R.H.Stherefore,\xa0{tex}\\sqrt { \\frac { \\sec \\theta - \\text{cosec} \\theta } { \\sec \\theta + \\text{cosec} \\theta } }{/tex}{tex}= \\frac { 1 } { \\sqrt { 7 } }{/tex}Hence proved.