Question

Show that sec? 0 - tane lies between 1 and 3 . seca e + tane

JEE/Engineering Exams

Maths

Solution

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( frac{sec ^{2} theta-tan theta}{sec ^{2} theta+tan theta}=y ) ( frac{1+tan ^{2} theta-tan theta}{=tan ^{2} theta+tan theta}=y ) ( frac{1+x^{2}-x}{1+x^{2}+x}=y quad[i t x=tan theta) ) ( =left{begin{array}{l}+x^{2}+x x^{2}-x+1=4left(x^{2}+x+1right)end{array}right. ) ( Rightarrow quad frac{x^{2}(1-y)-x(1+y)+(1-y)=0}{D geq 0} ) ( left[-(1+y)^{2}-4(1-y)^{2} geq 0right. ) ( 1+y^{2}+2 y=4left(1+y^{2}-2 yright)^{2} ) ( Rightarrow y^{2}-4 y^{2}+2 y+8 y+1-4 geq 0 ) ( 2 x^{0} ) ( y^{2}-4 y^{2}+2 y+8 y+1-4 geq 0 ) ( Rightarrow-3 y^{2}+10 y-3 geq 0 ) 7 ( 3 y^{2}-10 y+3 leq 0 ) ( Rightarrow(3 y-1)(y-3) leq 0 ) ( frac{1}{a} cdot x leq y leq 3 cdot begin{array}{l}x^{3} leq sec ^{2} theta-tan 0 leq 3 sec ^{2} theta+tan thetaend{array} )