Advertisements
Advertisements
प्रश्न
Prove that `tan^-1 (m/n) - tan^-1 ((m - n)/(m + n)) = pi/4`
उत्तर
LHS = `tan^-1 (m/n) - tan^-1 ((m - n)/(m + n))`
`= tan^-1 ((m/n - (m-n)/(m+n))/(1 + (m/n)((m-n)/(m+n))))`
`= tan^-1 (((m(m + n) - n(m - n))/(n(m+n)))/((n(m+n)+m(m-n))/(n(m+n))))`
`= tan^-1 ((m^2 + mn - nm + n^2)/(nm + n^2 + m^2 - mn))`
`= tan^-1 ((m^2 + n^2)/(m^2 + n^2))`
`= tan^-1 (1) = pi/4`
APPEARS IN
संबंधित प्रश्न
Find the value of the following:
`cos^(-1) (cos (13pi)/6)`
Prove the following:
`sin^-1(-1/2) + cos^-1(-sqrt(3)/2) = cos^-1(-1/2)`
The principal value of sin−1`(1/2)` is ______
Prove that `2 tan^-1 (3/4) = tan^-1(24/7)`
Evaluate `cos[pi/6 + cos^-1 (- sqrt(3)/2)]`
`sin^-1x + sin^-1 1/x + cos^-1x + cos^-1 1/x` = ______
In Δ ABC, with the usual notations, if sin B sin C = `"bc"/"a"^2`, then the triangle is ______.
What is the value of `sin^-1(sin (3pi)/4)`?
If cos–1 x > sin–1 x, then ______.
If sin–1x – cos–1x = `π/6`, then x = ______.
