Question Number 73789 by liki last updated on 15/Nov/19
Commented by liki last updated on 15/Nov/19
$$..{Sory}\:{sir};\:{Mr}\:{mind}\:{is}\:{power}\:{plz}\:{help}\: \\ $$$$\:\:{that}\:{qn}\:\mathrm{5}\:\left({a}\right),\left({b}\right),\left({c}\right)\:{and}\:\left({d}\right).\:{or}\:{anyone} \\ $$$$\:{to}\:{assist}\:{me}\:{this}\:{qn} \\ $$
Commented by mathmax by abdo last updated on 15/Nov/19
![let find tan(arcsinx) with x∈[−1,1] let arcsinx =t ⇒x =sint ⇒tan(arcsinx) =tant =((sint)/(cost)) =((sint)/(√(1−sin^2 t)))=(x/(√(1−x^2 ))) ⇒tan(arcsin((3/4)))=((3/4)/(√(1−((3/4))^2 ))) =(3/(4(√(1−(9/(16)))))) =(3/(√7))](Q73796.png)
$${let}\:{find}\:{tan}\left({arcsinx}\right)\:{with}\:{x}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${let}\:{arcsinx}\:={t}\:\Rightarrow{x}\:={sint}\:\Rightarrow{tan}\left({arcsinx}\right)\:={tant}\:=\frac{{sint}}{{cost}} \\ $$$$=\frac{{sint}}{\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}}=\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow{tan}\left({arcsin}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)=\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\sqrt{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}\sqrt{\mathrm{1}−\frac{\mathrm{9}}{\mathrm{16}}}}\:=\frac{\mathrm{3}}{\sqrt{\mathrm{7}}} \\ $$
Commented by mathmax by abdo last updated on 15/Nov/19
$$\left.{b}\right)\:{let}\:{arcsinx}\:={t}\:\:\:{with}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}\:\:\Rightarrow{x}={sint}\:{andcos}\left({arcsinx}\right) \\ $$$$={cos}\left({t}\right)=\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$
Commented by liki last updated on 16/Nov/19
$$...{Thanks}\:{sir}.. \\ $$
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)](Q73792.png)
$$\mathrm{tan}\left(\mathrm{x}\right)=\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{cos}\left(\mathrm{x}\right)}=\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}}, \\ $$$$\mathrm{tan}\left(\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)=\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\sqrt{\mathrm{1}−\frac{\mathrm{9}}{\mathrm{16}}}}=\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}}=\frac{\mathrm{3}}{\sqrt{\mathrm{7}}} \\ $$$$\left.\mathrm{b}\right)\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{sin}^{−} \left(\mathrm{t}\right),\mathrm{sin}\left(\mathrm{x}\right)=\mathrm{t} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{2}} \left(\mathrm{t}\right)+\mathrm{t}^{\mathrm{2}} =\mathrm{1}\Rightarrow\mathrm{cos}\left(\mathrm{x}\right)=\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} },\mathrm{cause}\:\mathrm{t}\in\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\:\:\mathrm{cos}\:\left(\mathrm{t}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{c}\left\{\mathrm{cot}\left(\theta\right)+\mathrm{cossec}\left(\theta\right)\right]^{\mathrm{2}} =\left(\frac{\mathrm{cos}\left(\theta\right)}{\mathrm{sin}\left(\theta\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\theta\right)}\right)^{\mathrm{2}} =\frac{\left(\mathrm{1}+\mathrm{cos}\left(\theta\right)\right)^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \left(\theta\right)}=\mathrm{A} \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\theta\right)=\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\theta\right)\right)=\left(\mathrm{1}−\mathrm{cos}\left(\theta\right)\right)\left(\mathrm{1}+\mathrm{cos}\left(\theta\right)\right) \\ $$$$\mathrm{A}=\frac{\left(\mathrm{1}+\mathrm{cos}\left(\theta\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{cos}\left(\theta\right)\right)\left(\mathrm{1}+\mathrm{cos}\left(\theta\right)\right)}=\frac{\mathrm{1}+\mathrm{cos}\left(\theta\right)}{\mathrm{1}−\mathrm{cos}\left(\theta\right)} \\ $$$$\left.\mathrm{d}\right)\mathrm{8sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2cos}\left(\mathrm{x}\right)−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\Rightarrow−\mathrm{8cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2cos}\left(\mathrm{x}\right)+\mathrm{3}=\mathrm{0} \\ $$$$\Delta=\mathrm{4}+\mathrm{96}=\mathrm{100} \\ $$$$\mathrm{cos}\left(\mathrm{x}\right)=\frac{−\mathrm{2}−\mathrm{10}}{\mathrm{2}.−\mathrm{8}}=\frac{\mathrm{12}}{\mathrm{16}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{or}\:\mathrm{cos}\left(\mathrm{x}\right)=\frac{−\mathrm{2}+\mathrm{10}}{\mathrm{2}.−\mathrm{8}}=\frac{\mathrm{8}}{−\mathrm{16}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow\mathrm{sin}\left(\mathrm{x}\right)=\underset{−} {+}\sqrt{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }=\underset{−} {+}\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{tg}\left(\mathrm{x}\right)=\underset{−} {+}\frac{\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$$\mathrm{cos}\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sin}\left(\mathrm{x}\right)=\underset{−} {+}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}\Rightarrow\mathrm{tg}\left(\mathrm{x}\right)=\underset{−} {+}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$
Commented by liki last updated on 15/Nov/19
$$\:{GOD}\:{bless}\:{you}\:{sir}.. \\ $$
Commented by mind is power last updated on 15/Nov/19
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$
Answered by Rio Michael last updated on 15/Nov/19
![Q5 b) cos[sin^(−1) (x)] =(√(1−x^2 )) LHS = cos[sin^(−1) x] let sin^(−1) x = u ⇒ x = sinu cosu = (√( 1−sin^2 u)) ⇒ cos[sin^(−1) x] = (√( 1−x^2 )) proved!](Q73801.png)
$$\left.{Q}\mathrm{5}\:{b}\right)\:{cos}\left[{sin}^{−\mathrm{1}} \left({x}\right)\right]\:=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${LHS}\:=\:{cos}\left[{sin}^{−\mathrm{1}} {x}\right]\: \\ $$$${let}\:{sin}^{−\mathrm{1}} {x}\:=\:{u}\:\Rightarrow\:{x}\:=\:{sinu} \\ $$$${cosu}\:=\:\sqrt{\:\mathrm{1}−{sin}^{\mathrm{2}} {u}} \\ $$$$\Rightarrow\:{cos}\left[{sin}^{−\mathrm{1}} {x}\right]\:=\:\sqrt{\:\mathrm{1}−{x}^{\mathrm{2}} }\:{proved}! \\ $$
Commented by liki last updated on 16/Nov/19
$$...{Thanks}\:{alot}\:{sir}.. \\ $$
Answered by Rio Michael last updated on 15/Nov/19
![tan[sin^(−1) ((3/4))] let sin^(−1) ((3/4)) = u ⇒ (3/4) = sinu ⇒tan[sin^(−1) ((3/4))] = α = tanu tanu = ((sinu)/(cosu)) = ((sinu)/(√(1−sin^2 u))) = ((3/4)/(√(1−(9/(16)))))= (3/(√7))](Q73803.png)
$${tan}\left[{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right]\: \\ $$$${let}\:{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\:=\:{u}\:\Rightarrow\:\frac{\mathrm{3}}{\mathrm{4}}\:=\:{sinu} \\ $$$$\Rightarrow{tan}\left[{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right]\:=\:\alpha\:=\:{tanu} \\ $$$${tanu}\:=\:\frac{{sinu}}{{cosu}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{{sinu}}{\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {u}}} \\ $$$$\:\:\:\:\:\:\:\:=\:\:\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\sqrt{\mathrm{1}−\frac{\mathrm{9}}{\mathrm{16}}}}=\:\frac{\mathrm{3}}{\sqrt{\mathrm{7}}} \\ $$$$ \\ $$