If tanA = (√3) , prove it: (√3) sinA cosA = (3/4)


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Question Number 126405 by amns last updated on 20/Dec/20

$If t a n A = \sqrt{3}, prove it : \sqrt{3} s i n A c o s A = \frac{3}{4}$
Commented by amns last updated on 20/Dec/20

$help me, plz . . .$
Commented by som(math1967) last updated on 20/Dec/20

$t a n A = \sqrt{3}$ $c o s A = \sqrt{\frac{1}{1 + {t a n}^{2} A}} = \frac{1}{2}$ $t a n A = \sqrt{3}$ $s i n A = t a n A c o s A = \frac{\sqrt{3}}{2}$ $\sqrt{3} s i n A c o s A$ $= \sqrt{3} \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{3}{4}$
	Answered by benjo_mathlover last updated on 20/Dec/20

	$i f A i n 1^{s t} - q u a d r a n t \Rightarrow \tan A = \sqrt{3}$ $\Rightarrow A = 60 ° \land \sqrt{3} \sin A . \cos A = \sqrt{3} \sin 60 ° . \cos 60 ° = \sqrt{3} \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{3}{4}$ $i f A i n 3^{r d} - q u a d r a n t \Rightarrow \tan A = \sqrt{3}$ $A = 240 ° t h e n \sqrt{3} \times (- \frac{\sqrt{3}}{2}) \times (- \frac{1}{2}) = \frac{3}{4}$
	Answered by TITA last updated on 20/Dec/20

	$\tan A = \sqrt{3} we show \sqrt{3} \sin Acos A = \frac{3}{4}$ $\sqrt{3} \cos A = \sin A \Rightarrow let t = \sqrt{3} \sin Acos A = \sin^{2} A$ $\Rightarrow t = \sin^{2} A = \frac{\tan^{2} A}{1 + \tan^{2} A} = \frac{3}{1 + 3} = \frac{3}{4}$
	Answered by MJS_new last updated on 20/Dec/20

	$t = \tan A \Rightarrow \sin A = \frac{t}{\sqrt{t^{2} + 1}} \land \cos A = \frac{1}{\sqrt{t^{2} + 1}}$ $\sqrt{3} \sin A \cos A = \frac{\sqrt{3} t}{t^{2} + 1} \overset{(t = \sqrt{3})}{=} \frac{3}{4}$
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