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Question Number 195578 by sonukgindia last updated on 05/Aug/23

Answered by Frix last updated on 05/Aug/23

sin α +cos α =((√3)/3) t=tan α ((t+1)/( (√(t^2 +1))))=((√3)/3) t=−((3−(√5))/2) ⇒ t^2 =((7−3(√5))/2)∧(1/t^2 )=((7+3(√5))/2)∧t^2 +(1/t^2 )=7

$$\mathrm{sin}\:\alpha\:+\mathrm{cos}\:\alpha\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${t}=\mathrm{tan}\:\alpha \\ $$$$\frac{{t}+\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${t}=−\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow \\ $$$${t}^{\mathrm{2}} =\frac{\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\wedge\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\wedge{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\mathrm{7} \\ $$

Answered by mr W last updated on 05/Aug/23

sin x+cos x=(1/( (√3))) 1+2 sin x cos x=((1/( (√3))))^2 =(1/3) ⇒sin x cos x=−(1/3) tan^2 x+cot^2 x =(tan x+(1/(tan x)))^2 −2 =((1/(sin x cos x)))^2 −2 =((1/(−(1/3))))^2 −2 =7

$$\mathrm{sin}\:{x}+\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:{x}+\mathrm{cot}^{\mathrm{2}} \:{x} \\ $$$$=\left(\mathrm{tan}\:{x}+\frac{\mathrm{1}}{\mathrm{tan}\:{x}}\right)^{\mathrm{2}} −\mathrm{2} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}\right)^{\mathrm{2}} −\mathrm{2} \\ $$$$=\left(\frac{\mathrm{1}}{−\frac{\mathrm{1}}{\mathrm{3}}}\right)^{\mathrm{2}} −\mathrm{2} \\ $$$$=\mathrm{7} \\ $$

Commented by Frix last updated on 05/Aug/23

tan x +(1/(tan x))=(1/(sin x cos x)) (1/((sin x cos x)^2 ))−2=(1/((−(1/3))^2 ))−2=9−2=7

$$\mathrm{tan}\:{x}\:+\frac{\mathrm{1}}{\mathrm{tan}\:{x}}=\frac{\mathrm{1}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\right)^{\mathrm{2}} }−\mathrm{2}=\frac{\mathrm{1}}{\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }−\mathrm{2}=\mathrm{9}−\mathrm{2}=\mathrm{7} \\ $$

Commented by mr W last updated on 05/Aug/23

yes, thanks!

$${yes},\:{thanks}! \\ $$

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