multi correct ans type Q Q y=tan^-1(2tan^2 y)-tan^-1(1/3tany),then tany is equal a -2 b 1 c 2/3 d 2 - Maths - Inverse Trigonometric Functions

Question

multi correct ans type Q
Q y=tan^-1(2tan^2 y)-tan^-1(1/3tany),then tany is equal
a -2 b 1 c 2/3 d 2
Q if sin[2cos^-1{cot(2tan^-1 x)}]=0,where x is greater than 0
,then
a x=1 b x=root(2)+1 c x=root(2)-1 d x=3

Himanshu · Accepted Answer

(1)

$y = \tan^{- 1} (2 \tan^{2} y) - \tan^{- 1} (\frac{1}{3 \tan y}) \Rightarrow y = \tan^{- 1} (\frac{2 \tan^{2} y - \frac{1}{3 \tan y}}{1 + 2 \tan^{2} y \times \frac{1}{3 \tan y}}) = \tan^{- 1} (\frac{6 \tan^{3} y - 1}{3 \tan y + 2 \tan^{2} y}) \Rightarrow \tan y = \frac{6 \tan^{3} y - 1}{3 \tan y + 2 \tan^{2} y} \Rightarrow 2 \tan^{3} y + 3 \tan^{2} y - 6 \tan^{3} y + 1 = 0 \Rightarrow 4 \tan^{3} y - 3 \tan^{2} y - 1 = 0 \Rightarrow 4 \tan^{3} y - 4 \tan^{2} y + \tan y - 1 = 0 \Rightarrow 4 \tan^{2} y (\tan y - 1) - (\tan y - 1) = 0 \Rightarrow (2 \tan y - 1) (2 \tan y + 1) (\tan y - 1) = 0 \Rightarrow \tan y = - \frac{1}{2} o r \frac{1}{2} o r 1$

(2)

$\sin [2 \cos^{- 1} (\cot (2 \tan^{- 1} x))] = 0 \tan^{- 1} x = θ \Rightarrow x = \tan θ \sin [2 \cos^{- 1} (\cot 2 θ)] = 0 \cot 2 θ = \cos y \sin [2 \cos^{- 1} (\cos y)] = \sin 2 y = 2 \cos y \sqrt{1 - \cos^{2} y} = 2 \cot 2 θ \sqrt{1 - \cot^{2} 2 θ} \tan θ = x \Rightarrow \tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{2 x}{1 - x^{2}} \sin [2 \cos^{- 1} (\cos y)] = \sin 2 y = 2 \cos y \sqrt{1 - \cos^{2} y} = 2 \cot 2 θ \sqrt{1 - \cot^{2} 2 θ} = \frac{2}{\tan^{2} 2 θ} \times \sqrt{\tan^{2} 2 θ - 1} \Rightarrow \frac{(1 - x^{2})^{2}}{2 x^{2}} \times \sqrt{\frac{4 x^{2}}{{(1 - x^{2})}^{2}} - 1} = 0 \Rightarrow \frac{1 - x^{2}}{2 x^{2}} \sqrt{6 x^{2} - x^{4} - 1} = 0 \frac{1 - x^{2}}{2 x^{2}} = 0 o r 6 x^{2} - x^{4} - 1 = 0 \Rightarrow x = 1 o r \Rightarrow x^{4} - 6 x^{2} + 1 = 0 \Rightarrow x^{2} = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2 \sqrt{2} = (\sqrt{2} \pm 1)^{2} x = \sqrt{2} \pm 1 o r 1$

multi correct ans.type Q. Q.y=tan^-1(2tan^2 y)-tan^-1(1/3tany),then tany is equal a.-2 b.1 c.2/3 d.2 Q.if sin[2cos^-1{cot(2tan^-1 x)}]=0,where x is greater than 0 ,then a.x=1 b.x=root(2)+1 c.x=root(2)-1 d.x=3

multi correct ans.type Q.

Q.y=tan^-1(2tan^2 y)-tan^-1(1/3tany),then tany is equal

a.-2 b.1 c.2/3 d.2

Q.if sin[2cos^-1{cot(2tan^-1 x)}]=0,where x is greater than 0 ,then

a.x=1 b.x=root(2)+1 c.x=root(2)-1 d.x=3