Educational

Value of 2tan⁻¹x + sin⁻¹(2x/(1+x²)) Question If \( x > 1 \), evaluate: \[ 2\tan^{-1}x + \sin^{-1}\left(\frac{2x}{1+x^2}\right) \] Solution Use identity: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \quad \text{(for } x \le 1\text{)} \] But for \( x > 1 \), principal value adjustment gives: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi – 2\tan^{-1}x \] Thus, \[ 2\tan^{-1}x + \left(\pi […]

Value of sin(2 tan⁻¹(0.75)) Question Evaluate: \[ \sin\left(2\tan^{-1}(0.75)\right) \] Solution Let \[ \theta = \tan^{-1}(0.75) = \tan^{-1}\left(\frac{3}{4}\right) \] Then, \[ \tan\theta = \frac{3}{4} \Rightarrow \text{Opposite} = 3,\ \text{Adjacent} = 4 \Rightarrow \text{Hypotenuse} = 5 \] \[ \sin\theta = \frac{3}{5}, \quad \cos\theta = \frac{4}{5} \] Now use identity: \[ \sin 2\theta = 2\sin\theta \cos\theta \] \[

Solve inverse trig equation for x Question If \[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \cos^{-1}\left(\frac{1-a^2}{1+a^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \] where \( a, x \in (0,1) \), find \( x \). Solution Let \[ a = \tan\theta \] Then, \[ \frac{2a}{1+a^2} = \sin 2\theta \quad,\quad \frac{1-a^2}{1+a^2} = \cos 2\theta \] So LHS becomes: \[ \sin^{-1}(\sin 2\theta) + \cos^{-1}(\cos 2\theta) \]

If tan⁻¹(cotθ) = 2θ, find θ Question If \[ \tan^{-1}(\cot\theta) = 2\theta \] Find \( \theta \). Solution Use identity: \[ \cot\theta = \tan\left(\frac{\pi}{2} – \theta\right) \] So, \[ \tan^{-1}(\cot\theta) = \tan^{-1}\left(\tan\left(\frac{\pi}{2} – \theta\right)\right) \] Using principal value: \[ \tan^{-1}(\tan x) = x \quad \text{if } x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Thus, \[ \tan^{-1}(\cot\theta) =

Value of cot(π/4 − 2cot⁻¹(3)) Question Evaluate: \[ \cot\left(\frac{\pi}{4} – 2\cot^{-1}(3)\right) \] Solution Let \[ \theta = \cot^{-1}(3) \Rightarrow \cot\theta = 3 \Rightarrow \tan\theta = \frac{1}{3} \] Find tan(2θ): \[ \tan 2\theta = \frac{2\tan\theta}{1 – \tan^2\theta} = \frac{2 \cdot \frac{1}{3}}{1 – \frac{1}{9}} = \frac{2/3}{8/9} = \frac{3}{4} \] Now use identity: \[ \cot\left(\frac{\pi}{4} – 2\theta\right) =

Value of sin(¼ sin⁻¹(√63/8)) Question Evaluate: \[ \sin\left(\frac{1}{4}\sin^{-1}\left(\frac{\sqrt{63}}{8}\right)\right) \] Solution Let \[ \theta = \sin^{-1}\left(\frac{\sqrt{63}}{8}\right) \Rightarrow \sin\theta = \frac{\sqrt{63}}{8} \] Then, \[ \cos\theta = \sqrt{1 – \frac{63}{64}} = \frac{1}{8} \] So, \[ \theta = \cos^{-1}\left(\frac{1}{8}\right) \] We need: \[ \sin\left(\frac{\theta}{4}\right) \] Use half-angle twice: \[ \cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}} = \sqrt{\frac{1+\frac{1}{8}}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]

Value of tan⁻¹(a/(b+c)) + tan⁻¹(b/(c+a)) Question In triangle ABC, if \( C = 90^\circ \), find: \[ \tan^{-1}\left(\frac{a}{b+c}\right) + \tan^{-1}\left(\frac{b}{c+a}\right) \] Solution In a right triangle at C: \[ c^2 = a^2 + b^2 \] Use identity: \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \] Let \[ x = \frac{a}{b+c}, \quad y = \frac{b}{c+a} \] Compute

If cos⁻¹x > sin⁻¹x, find range of x Question If \[ \cos^{-1}x > \sin^{-1}x \] Find the range of \( x \). Solution Use identity: \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \] So inequality becomes: \[ \cos^{-1}x > \frac{\pi}{2} – \cos^{-1}x \] \[ 2\cos^{-1}x > \frac{\pi}{2} \] \[ \cos^{-1}x > \frac{\pi}{4} \] Now apply cosine:

Solve tan⁻¹ equation Question If \[ \tan^{-1}\left(\frac{x+1}{x-1}\right) + \tan^{-1}\left(\frac{x-1}{x}\right) = \tan^{-1}(-7) \] Find \( x \). Solution Use identity: \[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \] Let \[ a = \frac{x+1}{x-1}, \quad b = \frac{x-1}{x} \] Compute numerator: \[ a + b = \frac{x+1}{x-1} + \frac{x-1}{x} = \frac{x(x+1) + (x-1)^2}{x(x-1)} \] \[ = \frac{x^2 +

If 4cos⁻¹x + sin⁻¹x = π, find x Question If \[ 4\cos^{-1}x + \sin^{-1}x = \pi \] Find \( x \). Solution Use identity: \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \Rightarrow \sin^{-1}x = \frac{\pi}{2} – \cos^{-1}x \] Substitute: \[ 4\cos^{-1}x + \left(\frac{\pi}{2} – \cos^{-1}x\right) = \pi \] \[ 3\cos^{-1}x + \frac{\pi}{2} = \pi \] \[