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प्रश्न
If \[\cos^{- 1} x > \sin^{- 1} x\], then
विकल्प
\[\frac{1}{\sqrt{2}} < x \leq 1\]
\[0 \leq x < \frac{1}{\sqrt{2}}\]
\[- 1 \leq x < \frac{1}{\sqrt{2}}\]
x > 0
उत्तर
\[\cos^{- 1} x > \sin^{- 1} x\]
\[ \Rightarrow \cos^{- 1} x > \frac{\pi}{2} - \cos^{- 1} x\]
\[ \Rightarrow 2 \cos^{- 1} x > \frac{\pi}{2}\]
\[ \Rightarrow \cos^{- 1} x > \frac{\pi}{4}\]
\[ \Rightarrow x > \cos\frac{\pi}{4}\]
\[ \Rightarrow x > \frac{1}{\sqrt{2}}\]
We know that the maximum value of cosine fuction is 1.
\[\therefore \frac{1}{\sqrt{2}} < x \leq 1\]
Hence, the correct answer is option(a).
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