ত্রিকোণমিতিক অনুপাত
sin−1x+sin−11x+cos−1x+cos−11x,x∉±1\sin^{-1}x+\sin^{-1}\dfrac{1}{x}+\cos^{-1}x+\cos^{-1}\dfrac{1}{x}, x\notin \pm 1sin−1x+sin−1x1+cos−1x+cos−1x1,x∈/±1 is equal to?
π\piπ
π2\dfrac{\pi}{2}2π
3π2\dfrac{3\pi}{2}23π
None of these
Solution
let's proof that sin−1x+cos−1x=π2 \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} sin−1x+cos−1x=2π Let sin−1x=0⇒x=sinθ=cos(π2−θ) \sin ^{-1} x=0 \Rightarrow x=\sin \theta=\cos \left(\frac{\pi}{2}-\theta\right) sin−1x=0⇒x=sinθ=cos(2π−θ) now, cos−1x=π2−θ=π2−sin−1x \cos ^{-1} x=\frac{\pi}{2}-\theta=\frac{\pi}{2}-\sin ^{-1} x cos−1x=2π−θ=2π−sin−1x.
∴sin−1x+cos−1x=π2. \therefore \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \text {. } ∴sin−1x+cos−1x=2π.
So, sin−1x+cos−1x+sin−11x+cos−11x \sin ^{-1} x+\cos ^{-1} x+\sin ^{-1} \frac{1}{x}+\cos ^{-1} \frac{1}{x} sin−1x+cos−1x+sin−1x1+cos−1x1
=π2+π2=π \begin{array}{l} =\frac{\pi}{2}+\frac{\pi}{2} \\ =\pi \end{array} =2π+2π=π
where x∈(−1,1) x \in(-1,1) x∈(−1,1).
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