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
`sin^-1 1/2-2sin^-1 1/sqrt2`
In ΔABC, if a = 18, b = 24, c = 30 then find the values of sinA
Evaluate the following:
`"cosec"^-1(-sqrt(2)) + cot^-1(sqrt(3))`
Prove the following:
`2tan^-1(1/3) = tan^-1(3/4)`
Prove that:
2 tan-1 (x) = `sin^-1 ((2x)/(1 + x^2))`
lf `sqrt3costheta + sintheta = sqrt2`, then the general value of θ is ______
`"sin"^-1 (-1/2)`
Which of the following functions is inverse of itself?
If –1 ≤ x ≤ 1, the prove that sin–1 x + cos–1 x = `π/2`
The value of `tan(cos^-1 4/5 + tan^-1 2/3)` is ______.
