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Question Number 182200 by Shrinava last updated on 05/Dec/22

1.   lim_(a→−x)  ((sin(x^2  − a^2 ))/(x^3  + a^3 )) = ?    2.   cot 80° (tan 10° + 2 tg 70°) = ?

$$\mathrm{1}.\:\:\:\underset{\boldsymbol{\mathrm{a}}\rightarrow−\boldsymbol{\mathrm{x}}} {\mathrm{lim}}\:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{a}^{\mathrm{3}} }\:=\:? \\ $$$$ \\ $$$$\mathrm{2}.\:\:\:\mathrm{cot}\:\mathrm{80}°\:\left(\mathrm{tan}\:\mathrm{10}°\:+\:\mathrm{2}\:\mathrm{tg}\:\mathrm{70}°\right)\:=\:? \\ $$

Answered by cortano1 last updated on 06/Dec/22

1. lim_(a→−x)  ((sin (x+a)(x−a))/((x+a)(x^2 −ax+a^2 )))   = ((2x)/(3x^2 )) = (2/(3x))

$$\mathrm{1}.\:\underset{\mathrm{a}\rightarrow−\mathrm{x}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{x}+\mathrm{a}\right)\left(\mathrm{x}−\mathrm{a}\right)}{\left(\mathrm{x}+\mathrm{a}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{ax}+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$\:=\:\frac{\mathrm{2x}}{\mathrm{3x}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}}{\mathrm{3x}} \\ $$

Answered by som(math1967) last updated on 06/Dec/22

2. tan80=((tan70+tan10)/(1−tan70tan10))   ⇒tan80−tan70.tan10.tan80        =tan70+tan10  ⇒tan80−tan70.tan10.cot10      =tan70+tan10  ⇒tan80−tan10=2tan70  now cot80(tan10+2tan70)  =cot80(tan10+tan80−tan10)  =cot80×tan80=1 ans

$$\mathrm{2}.\:{tan}\mathrm{80}=\frac{{tan}\mathrm{70}+{tan}\mathrm{10}}{\mathrm{1}−{tan}\mathrm{70}{tan}\mathrm{10}} \\ $$$$\:\Rightarrow{tan}\mathrm{80}−{tan}\mathrm{70}.{tan}\mathrm{10}.{tan}\mathrm{80} \\ $$$$\:\:\:\:\:\:={tan}\mathrm{70}+{tan}\mathrm{10} \\ $$$$\Rightarrow{tan}\mathrm{80}−{tan}\mathrm{70}.{tan}\mathrm{10}.{cot}\mathrm{10} \\ $$$$\:\:\:\:={tan}\mathrm{70}+{tan}\mathrm{10} \\ $$$$\Rightarrow{tan}\mathrm{80}−{tan}\mathrm{10}=\mathrm{2}{tan}\mathrm{70} \\ $$$${now}\:{cot}\mathrm{80}\left({tan}\mathrm{10}+\mathrm{2}{tan}\mathrm{70}\right) \\ $$$$={cot}\mathrm{80}\left({tan}\mathrm{10}+{tan}\mathrm{80}−{tan}\mathrm{10}\right) \\ $$$$={cot}\mathrm{80}×{tan}\mathrm{80}=\mathrm{1}\:{ans} \\ $$$$ \\ $$

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