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`        solve for xsin[cot^-1(x+1)]=cos[tan^-1(x)]when i write cot inverse x as pi by 2 – tan inverse x. x gets cancelled`
3 years ago

```							cot^-1(x+1) = sin^-1(1/sqrt(1+(1+x)²)tan^-1(x) = cos^-1(1/sqrt(x²+1))hence,  sin(cot^-1(x+1))    = cos(tan^-1(x))             sin(sin^-1(1/sqrt(1+(1+x)²))=cos(cos^-1(1/sqrt(x²+1)))             1/sqrt(1+(1+x)²=1/sqrt(x²+1)      ==> 1+(1+x)² = x² + 1     ==> 1 + 1 + x² + 2x = x² + 1     ==>  2x = -1    ==> x = – 1/2
```
one year ago
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