Prove: `sin^(-1)(1/sqrt5)+cot^(-1)3=pi/4`

1.	Prove: `sin^(-1)(1/sqrt5)+cot^(-1)3=pi/4`
Answer» Let `cot^-1 3 = theta`. Then, `cot theta = 3` `=>costheta/sintheta = 3 =>cos^2theta/sin^2theta = 9` `=>(1-sin^2theta)/sin^2theta = 9 =>sintheta = 1/sqrt10` `=> theta = sin^-1(1/sqrt10)` `:. L.H.S. = sin^-1(1/sqrt5) +sin^-1(1/sqrt10)` We know, `sin^-1x+sin^-1y = sin^-1(xsqrt(1-y^2)+ysqrt(1-x^2))` So, our expression becomes, `= sin^-1(1/sqrt5sqrt(1-1/10)+sqrt(1-1/5)1/sqrt10)` `=sin^-1(3/sqrt50+2/sqrt50)` `=sin^-1(5/(5sqrt2))` `=sin^-1(1/sqrt2)` `=pi/4 = R.H.S.`

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Prove: `sin^(-1)(1/sqrt5)+cot^(-1)3=pi/4`

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