Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

$\left(\text{P}\right)-4\text{;}\left(\text{Q}\right)-3\text{;}\left(\text{R}\right)-2\text{or}4\text{;}\left(\text{S}\right)-1$
$\left(\text{P}\right){\left(\frac{1}{y^2}{\left(\frac{\text{cos}\left({\text{tan}}^{-1}y\right)+y\text{sin}\left({\text{tan}}^{-1}y\right)}{\text{cot}\left({\text{sin}}^{-1}y\right)+\text{tan}\left({\text{sin}}^{-1}y\right)}\right)}^2+y_4\right)}^{1/2}$
$={\left[\frac{1}{y^2}{\left[\frac{\left(\frac{1}{\sqrt{1+y^2}}+\frac{y\text{.}y}{\sqrt{1+y^2}}\right)}{\left(\frac{\sqrt{1-y^2}}{y}+\frac{y}{\sqrt{1-y^2}}\right)}\right]}^2+y^4\right]}^{1/2}$
$=\frac{{\text{y}}^2}{{\text{y}}^2}\left(1-{\text{y}}^2\right)+{\text{y}}^4$
$=1$

4
$\left(\text{Q}\right)\text{cos}x+\text{cos}y=-\text{cos}z$
$\text{sin}x+\text{sin}y=-\text{sin}z\text{Square and add}$
$2+2\text{cos}\left(x-y\right)=1$
$\Rightarrow \text{cos}\left(x-y\right)=-1/2$
$\Rightarrow 2{\text{cos}}^2\left(\frac{x-y}{2}\right)-1=-1/2\text{,}\Rightarrow \text{cos}\left(\frac{x-y}{2}\right)=1/2$
3
$\left(\text{R}\right)\text{cos}2x\left(\text{cos}\left(\frac{\pi }{4}-x\right)-\text{cos}\left(\frac{\pi }{4}+x\right)\right)+2{\text{sin}}^{2}x=2\text{sin}x\text{cos}x$
$\text{cos}2x\left(\sqrt{2}sinx\right)+2{\text{sin}}^{2}x=2\text{sin}x\text{cos}x$
$\sqrt{2}\text{sin}x\left[\text{cos}2x+\sqrt{2}\text{sin}x-\sqrt{2}\text{cos}x\right]=0$
Either sin x = 0 OR${\text{cos}}^{2}x-{\text{sin}}^{2}x=\sqrt{2}\left(\text{cos}x-\text{sin}x\right)$
sec x = 1    OR    cos x = sin x
$\Rightarrow \text{sec}x=\sqrt{2}$
2 OR 4
$\left(\text{S}\right)\text{cot}\left({\text{sin}}^{-1}\sqrt{1-x^2}\right)=\text{sin}\left({\text{tan}}^{-1}\left(x\sqrt{6}\right)\right)$

$\frac{x}{\sqrt{1-x^2}}=\frac{x\sqrt{6}}{\sqrt{1+6x^2}}$
$\Rightarrow 1+6x^2=6-6x^2$
$\Rightarrow 12x^2=5$
$\Rightarrow x=\sqrt{\frac{5}{12}}==\frac{1}{2}\sqrt{\frac{5}{3}}$
1