The total number of real solutions of the equation $θ = \tan^{- 1} (2 \tan θ) - \frac{1}{2} \sin^{- 1} (\frac{6 \tan θ}{9 + \tan^{2} θ})$ is

(Here, the inverse trigonometric functions $\sin^{- 1} x$ and $\tan^{- 1} x$ assume values in $[- \frac{π}{2}, \frac{π}{2}]$ and $(- \frac{π}{2}, \frac{π}{2})$ , respectively.) [2025]

Question

The total number of real solutions of the equation $θ = \tan^{- 1} (2 \tan θ) - \frac{1}{2} \sin^{- 1} (\frac{6 \tan θ}{9 + \tan^{2} θ})$ is

(Here, the inverse trigonometric functions $\sin^{- 1} x$ and $\tan^{- 1} x$ assume values in $[- \frac{π}{2}, \frac{π}{2}]$ and $(- \frac{π}{2}, \frac{π}{2})$ , respectively.) [2025]

Smart Achievers · Accepted Answer

(3)

$θ = \tan^{- 1} (2 \tan θ) - \frac{1}{2} \sin^{- 1} (\frac{6 \tan θ}{9 + \tan^{2} θ})$

Let $\tan θ = 1$ and taking tangent on both sides,

$t = \frac{2 t - \tan (\frac{1}{2} \sin^{- 1} (\frac{6 t}{9 + t^{2}}))}{1 + 2 t \tan (\frac{1}{2} \sin^{- 1} (\frac{6 t}{9 + t^{2}}))}$ $\dots (i)$

Let $\sin^{- 1} (\frac{6 t}{9 + t^{2}}) = α$

$\sin α = \frac{6 t}{9 + t^{2}}$

$\frac{2 \tan \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} = \frac{2 (\frac{t}{3})}{1 + {(\frac{t}{3})}^{2}} = \frac{2 (\frac{3}{t})}{{(\frac{3}{t})}^{2} + 1}$

So, $\tan \frac{α}{2} = \frac{3}{t} or \frac{t}{3}$

$t = \frac{2 t - \tan (\frac{α}{2})}{1 + 2 t . \tan (\frac{α}{2})}$ [From (i)]

Case-I: If $\tan \frac{α}{2} = \frac{3}{t}$

$\Rightarrow t = \frac{2 t - \frac{3}{t}}{1 + 6}$ $\Rightarrow 7 t^{2} = 2 t^{2} - 3$

$\Rightarrow 5 t^{2} = - 3,$

which is not possible.

Case-II: If $\tan \frac{α}{2} = \frac{t}{3},$

$\Rightarrow t = \frac{2 t - \frac{t}{3}}{1 + \frac{2 t^{2}}{3}}$ $\Rightarrow t (3 + 2 t^{2}) = 5 t$

$\Rightarrow 3 + 2 t^{2} = 5$

$\Rightarrow t = 0, 1, - 1 = \tan θ$

$\Rightarrow θ = 0, \frac{π}{4}, - \frac{π}{4}$

Therefore, the solution are ${\frac{- π}{4}, 0, \frac{π}{4}}$

Solution

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Solution

Q. The total number of real solutions of the equation θ=tan-1(2tanθ)-12sin-1(6tanθ9+tan2θ) is (Here, the inverse trigonometric functions sin-1x and tan-1x assume values in [-π2,π2] and (-π2,π2), respectively.) [2025]

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