tg-15-ctg-5-

Question Number 227312 by hardmath last updated on 13/Jan/26

$$\mathrm{tg}\left(\mathrm{15}\right)\:+\:\mathrm{ctg}\left(\mathrm{5}\right)\:=\:? \\ $$

Answered by Kassista last updated on 13/Jan/26

tg(3θ) = ((3tg(θ)−tg^3 (θ))/(1−3tg^2 (θ))) ∴ tg(15)=((3tg(5)−tg^3 (5))/(1−3tg^2 (5))) ⇒ tg(15)+cotg(5) = ((3tg(5)−tg^3 (5))/(1−3tg^2 (5)))+(1/(tg(5))) = ((3tg^2 (5)−tg^4 (5)+1−3tg^2 (5))/(tg(5)−3tg^3 (5))) = ((1−tg^4 (5))/(tg(5)(1−3tg^2 (5)))) (((1+tg^2 (5))(1−tg^2 (5)))/(tg(5)(1−3tg^2 (5)))) = ((sec^2 (5)(1+tg(5))(1−tg(5)))/(tg(5)(1−3tg^2 (5)))) = ((2(1+tg(5))(1−tg(5)))/(sin(10)(1−3tg^2 (5)))) ≈ 11.698001

$$ \\ $$$${tg}\left(\mathrm{3}\theta\right)\:=\:\frac{\mathrm{3}{tg}\left(\theta\right)−{tg}^{\mathrm{3}} \left(\theta\right)}{\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\theta\right)}\:\therefore\:{tg}\left(\mathrm{15}\right)=\frac{\mathrm{3}{tg}\left(\mathrm{5}\right)−{tg}^{\mathrm{3}} \left(\mathrm{5}\right)}{\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\mathrm{5}\right)} \\ $$$$ \\ $$$$\Rightarrow\:{tg}\left(\mathrm{15}\right)+{cotg}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{3}{tg}\left(\mathrm{5}\right)−{tg}^{\mathrm{3}} \left(\mathrm{5}\right)}{\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\mathrm{5}\right)}+\frac{\mathrm{1}}{{tg}\left(\mathrm{5}\right)} \\ $$$$=\:\frac{\mathrm{3}{tg}^{\mathrm{2}} \left(\mathrm{5}\right)−{tg}^{\mathrm{4}} \left(\mathrm{5}\right)+\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\mathrm{5}\right)}{{tg}\left(\mathrm{5}\right)−\mathrm{3}{tg}^{\mathrm{3}} \left(\mathrm{5}\right)}\:=\:\frac{\mathrm{1}−{tg}^{\mathrm{4}} \left(\mathrm{5}\right)}{{tg}\left(\mathrm{5}\right)\left(\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\mathrm{5}\right)\right)} \\ $$$$ \\ $$$$\frac{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left(\mathrm{5}\right)\right)\left(\mathrm{1}−{tg}^{\mathrm{2}} \left(\mathrm{5}\right)\right)}{{tg}\left(\mathrm{5}\right)\left(\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\mathrm{5}\right)\right)}\:=\:\frac{{sec}^{\mathrm{2}} \left(\mathrm{5}\right)\left(\mathrm{1}+{tg}\left(\mathrm{5}\right)\right)\left(\mathrm{1}−{tg}\left(\mathrm{5}\right)\right)}{{tg}\left(\mathrm{5}\right)\left(\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\mathrm{5}\right)\right)} \\ $$$$ \\ $$$$=\:\frac{\mathrm{2}\left(\mathrm{1}+{tg}\left(\mathrm{5}\right)\right)\left(\mathrm{1}−{tg}\left(\mathrm{5}\right)\right)}{{sin}\left(\mathrm{10}\right)\left(\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\mathrm{5}\right)\right)}\:\approx\:\mathrm{11}.\mathrm{698001} \\ $$$$ \\ $$

Answered by Frix last updated on 13/Jan/26

$$\mathrm{2}−\sqrt{\mathrm{3}}+\mathrm{cot}\:\mathrm{5}° \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply