Question

Cos^-1 x+ sin^-1/√5 = π/4

Answer 1

Answer:

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