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Question Number 73789 by liki last updated on 15/Nov/19

Commented by liki last updated on 15/Nov/19

$. . S o r y s i r; M r m i n d i s p o w e r p l z h e l p$ $t h a t q n 5 (a), (b), (c) a n d (d) . o r a n y o n e$ $t o a s s i s t m e t h i s q n$
Commented by mathmax by abdo last updated on 15/Nov/19

$l e t f i n d t a n (a r c s i n x) w i t h x \in [- 1, 1]$ $l e t a r c s i n x = t \Rightarrow x = s i n t \Rightarrow t a n (a r c s i n x) = t a n t = \frac{s i n t}{c o s t}$ $= \frac{s i n t}{\sqrt{1 - {s i n}^{2} t}} = \frac{x}{\sqrt{1 - x^{2}}} \Rightarrow t a n (a r c s i n (\frac{3}{4})) = \frac{\frac{3}{4}}{\sqrt{1 - {(\frac{3}{4})}^{2}}}$ $= \frac{3}{4 \sqrt{1 - \frac{9}{16}}} = \frac{3}{\sqrt{7}}$
Commented by mathmax by abdo last updated on 15/Nov/19

$b) l e t a r c s i n x = t w i t h - 1 ⩽ x ⩽ 1 \Rightarrow x = s i n t a n d c o s (a r c s i n x)$ $= c o s (t) = \sqrt{1 - {s i n}^{2} t} = \sqrt{1 - x^{2}}$
Commented by liki last updated on 16/Nov/19

$. . . T h a n k s s i r . .$
	Answered by mind is power last updated on 15/Nov/19
	$tan(x)=((sin(x))/(cos(x)))=((sin(x))/(√(1−sin^2 (x)))), tan(sin^(−1) ((3/4)))=((3/4)/(√(1−(9/(16)))))=((3/4)/((√7)/4))=(3/(√7)) b)cos^2 (x)+sin^2 (x)=1⇒x=sin^− (t),sin(x)=t ⇒cos^2 (t)+t^2 =1⇒cos(x)=(√(1−t^2 )),cause t∈[−(π/2),(π/2)] cos (t)≥0 c{cot(θ)+cossec(θ)]^2 =(((cos(θ))/(sin(θ)))+(1/(sin(θ))))^2 =(((1+cos(θ))^2 )/(sin^2 (θ)))=A sin^2 (θ)=(1−cos^2 (θ))=(1−cos(θ))(1+cos(θ)) A=(((1+cos(θ))^2 )/((1−cos(θ))(1+cos(θ))))=((1+cos(θ))/(1−cos(θ))) d)8sin^2 (x)+2cos(x)−5=0 sin^2 (x)=1−cos^2 (x)⇒−8cos^2 (x)+2cos(x)+3=0 Δ=4+96=100 cos(x)=((−2−10)/(2.−8))=((12)/(16))=(3/4) or cos(x)=((−2+10)/(2.−8))=(8/(−16))=−(1/2) cos(x)=(3/4)⇒sin(x)=+_− (√(1−((3/4))^2 ))=+_− ((√7)/4) ⇒tg(x)=+_− ((√7)/3) cos(x)=−(1/2)⇒sin(x)=+_− (√(3/4))⇒tg(x)=+_− (√3)$
	$\tan (x) = \frac{\sin (x)}{\cos (x)} = \frac{\sin (x)}{\sqrt{1 - \sin^{2} (x)}},$ $\tan (\sin^{- 1} (\frac{3}{4})) = \frac{\frac{3}{4}}{\sqrt{1 - \frac{9}{16}}} = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}}$ $b) \cos^{2} (x) + \sin^{2} (x) = 1 \Rightarrow x = \sin^{-} (t), \sin (x) = t$ $\Rightarrow \cos^{2} (t) + t^{2} = 1 \Rightarrow \cos (x) = \sqrt{1 - t^{2}}, cause t \in [- \frac{π}{2}, \frac{π}{2}] \cos (t) ⩾ 0$ $c {{\cot (θ) + cossec (θ)]}^{2} = {(\frac{\cos (θ)}{\sin (θ)} + \frac{1}{\sin (θ)})}^{2} = \frac{{(1 + \cos (θ))}^{2}}{\sin^{2} (θ)} = A$ $\sin^{2} (θ) = (1 - \cos^{2} (θ)) = (1 - \cos (θ)) (1 + \cos (θ))$ $A = \frac{{(1 + \cos (θ))}^{2}}{(1 - \cos (θ)) (1 + \cos (θ))} = \frac{1 + \cos (θ)}{1 - \cos (θ)}$ $d) {8 \sin}^{2} (x) + 2 \cos (x) - 5 = 0$ $\sin^{2} (x) = 1 - \cos^{2} (x) \Rightarrow - {8 \cos}^{2} (x) + 2 \cos (x) + 3 = 0$ $Δ = 4 + 96 = 100$ $\cos (x) = \frac{- 2 - 10}{2 . - 8} = \frac{12}{16} = \frac{3}{4}$ $or \cos (x) = \frac{- 2 + 10}{2 . - 8} = \frac{8}{- 16} = - \frac{1}{2}$ $\cos (x) = \frac{3}{4} \Rightarrow \sin (x) = \underline{+} \sqrt{1 - {(\frac{3}{4})}^{2}} = \underline{+} \frac{\sqrt{7}}{4}$ $\Rightarrow tg (x) = \underline{+} \frac{\sqrt{7}}{3}$ $\cos (x) = - \frac{1}{2} \Rightarrow \sin (x) = \underline{+} \sqrt{\frac{3}{4}} \Rightarrow tg (x) = \underline{+} \sqrt{3}$
	Commented by liki last updated on 15/Nov/19

	$G O D b l e s s y o u s i r . .$
	Commented by mind is power last updated on 15/Nov/19

	$y^{'} re welcom$
	Answered by Rio Michael last updated on 15/Nov/19

	$Q 5 b) c o s [{s i n}^{- 1} (x)] = \sqrt{1 - x^{2}}$ $L H S = c o s [{s i n}^{- 1} x]$ $l e t {s i n}^{- 1} x = u \Rightarrow x = s i n u$ $c o s u = \sqrt{1 - {s i n}^{2} u}$ $\Rightarrow c o s [{s i n}^{- 1} x] = \sqrt{1 - x^{2}} p r o v e d!$
	Commented by liki last updated on 16/Nov/19

	$. . . T h a n k s a l o t s i r . .$
	Answered by Rio Michael last updated on 15/Nov/19

	$t a n [{s i n}^{- 1} (\frac{3}{4})]$ $l e t {s i n}^{- 1} (\frac{3}{4}) = u \Rightarrow \frac{3}{4} = s i n u$ $\Rightarrow t a n [{s i n}^{- 1} (\frac{3}{4})] = α = t a n u$ $t a n u = \frac{s i n u}{c o s u}$ $= \frac{s i n u}{\sqrt{1 - {s i n}^{2} u}}$ $= \frac{\frac{3}{4}}{\sqrt{1 - \frac{9}{16}}} = \frac{3}{\sqrt{7}}$
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