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Question : 24 of 160
Marks:
+1,
-0
Solution:
‌tan−1(1)+‌cos−1(x2)‌−tan−1(‌| √1+x2+√1−x2 |
| √1+x2−√1−x2 |
)=0‌∵tan−1(1)=π∕4‌ and let ‌θ=cos−1(x2)‌⇒x2=cos‌θ‌‌ and ‌1+x2=1+cos‌x,1−x2=1−cos‌x‌=‌+‌−tan−1(‌| √1+cos‌θ+√1−cos‌θ |
| √1+cos‌θ−√1−cos‌θ |
)‌=‌+‌−tan−1(‌| √2‌cos‌θ∕2+√2sin‌θ∕2 |
| √2‌cos‌θ∕2−√2sin‌θ∕2 |
)‌=‌+‌−tan−1(‌| cos‌θ∕2+sin‌θ∕2 |
| cos‌θ∕2−sin‌θ∕2 |
)‌∵‌| cos‌x+sin‌x |
| cos‌x−sin‌x |
=tan(‌+x)‌=‌+‌−tan−1(tan(‌+‌))‌=‌+‌−‌−‌=0Hence, the identify is true for all
x Infinite many solutions.
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