tan cote _=1 + sec 0 cosec e (111) ...
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tan cote _=1 + sec 0 cosec e (111) 1-cote 1- tan [Hint : Write the expression in terms of sin 0 and cose]

11th - 12th Class
Maths
Solution
141
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(iii) ( frac{tan theta}{1-cot theta}+frac{cot theta}{1-tan theta}=1+sec theta operatorname{cosec} theta ) LHS ( =frac{tan theta}{1-cot theta}+frac{cot theta}{1-tan theta} ldots ldots ldots ) Equation (1) We know that the trigonometric functions, ( tan (x)=frac{sin (x)}{cos (x)} ) ( cot (x)=frac{cos (x)}{sin (x)}=frac{1}{tan (x)} ) By substituting the above function in Equation ( 1 ). ( =frac{frac{sin theta}{cos theta}}{1-frac{cos theta}{sin theta}}+frac{frac{cos theta}{sin theta}}{1-frac{cos theta}{sin theta}} ) ( =frac{frac{sin theta}{cos theta}}{frac{sin theta-cos theta}{sin theta}}+frac{frac{cos theta}{sin theta}}{frac{cos theta-sin theta}{cos theta}} quad ) (By taking LCM and Common denominators) ( =frac{sin ^{2} theta}{cos theta(sin theta-cos theta)}+frac{cos ^{2} theta}{sin theta(sin theta-cos theta)} ) Taking ( frac{1}{(sin theta-cos theta)} ) as common ( =frac{1}{(sin theta-cos theta)}left[frac{sin ^{2} theta}{cos theta}-frac{cos ^{2} theta}{sin theta}right] ) ( =frac{1}{(sin theta-cos theta)}left[frac{sin ^{3} theta-cos ^{3} theta}{sin theta cos theta}right] ) Using ( a^{3}-b^{3}=(a-b)left(a^{2}+a b+b^{2}right) ) wore ( =frac{1}{(sin theta-cos theta)}left[frac{(sin theta-cos theta)left(sin ^{2} theta+cos ^{2} theta+sin theta cos thetaright)}{sin theta cos theta}right] ) ( left.=frac{(1+sin theta cos theta)}{(sin theta cos theta)} quad text { (By Identity } sin ^{2} A+cos ^{2} A=1right) ) ( =1+sec theta operatorname{cosec} theta ) ( =mathrm{R} mathrm{H} . mathrm{S} )
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