Question

prove that:- cot inverse x + sin inverse 1/√5= π/ 4​

Answer 1

we have to prove that : $cot^{-1}3+sin^{-1}\frac{1}{\sqrt{5}}=\frac{\pi}{4}$

Let $cot^{-1}3=A$

cotA = 3 = b/p

so, h = √(b² + p²) = √(3² + 1²) = √(10)

so, sinA = p/h = 1/√10

or, $A=sin^{-1}\frac{1}{\sqrt{10}}$

hence, LHS = $sin^{-1}\frac{1}{\sqrt{10}}+sin^{-1}\frac{1}{\sqrt{10}}$

we know, $sin^{-1}x+sin^{-1}y=sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})$

$\textbf{so},sin^{-1}\frac{1}{\sqrt{10}}+sin^{-1}\frac{1}{\sqrt{10}}=sin^{-1}\left(\frac{1}{\sqrt{10}}\times\sqrt{1-\left(\frac{1}{\sqrt{5}}\right)^2}+\frac{1}{\sqrt{5}}\sqrt{1-\left(\frac{1}{\sqrt{10}}\right)^2}\right)$

$=sin^{-1}\left(\frac{1}{\sqrt{10}}\times\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{10}}\times\frac{3}{\sqrt{5}}\right)$

$=sin^{-1}\left(\frac{5}{\sqrt{50}}\right)$

$=sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$\frac{\pi}{4}=RHS$