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Question-82176




Question Number 82176 by oyemi kemewari last updated on 19/Feb/20
Commented by jagoll last updated on 19/Feb/20
(1b) (√(32−2(√(135)))) = (√(27+5−2(√(27×5))))  = (√(27))−(√5) = 3(√3)−(√5)
$$\left(\mathrm{1}{b}\right)\:\sqrt{\mathrm{32}−\mathrm{2}\sqrt{\mathrm{135}}}\:=\:\sqrt{\mathrm{27}+\mathrm{5}−\mathrm{2}\sqrt{\mathrm{27}×\mathrm{5}}} \\ $$$$=\:\sqrt{\mathrm{27}}−\sqrt{\mathrm{5}}\:=\:\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}} \\ $$
Commented by jagoll last updated on 19/Feb/20
(2b) α×(α+3) = 28  α^2 +3α−28=0  (α+7)(α−4)=0  (i) k = 2α+3 = −11  (ii) k = 2α+3 = 11
$$\left(\mathrm{2}{b}\right)\:\alpha×\left(\alpha+\mathrm{3}\right)\:=\:\mathrm{28} \\ $$$$\alpha^{\mathrm{2}} +\mathrm{3}\alpha−\mathrm{28}=\mathrm{0} \\ $$$$\left(\alpha+\mathrm{7}\right)\left(\alpha−\mathrm{4}\right)=\mathrm{0} \\ $$$$\left({i}\right)\:{k}\:=\:\mathrm{2}\alpha+\mathrm{3}\:=\:−\mathrm{11} \\ $$$$\left({ii}\right)\:{k}\:=\:\mathrm{2}\alpha+\mathrm{3}\:=\:\mathrm{11} \\ $$
Commented by john santu last updated on 19/Feb/20
((cos x)/(1+sin x)) + ((1+sin x)/(cos x)) =   ((cos^2 x+1+2sin x+sin^2 x)/(cos x(1+sin x))) =?  ((2(1+sin x))/(cos x(1+sin x))) = (2/(cos x)) = 2sec x
$$\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:+\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\:=\: \\ $$$$\frac{\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{1}+\mathrm{2sin}\:{x}+\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:{x}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}\:=? \\ $$$$\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}{\mathrm{cos}\:{x}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{cos}\:{x}}\:=\:\mathrm{2sec}\:{x} \\ $$

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