Advertisements
Advertisements
Question
Simplify: `tan^-1 x/y - tan^-1 (x - y)/(x + y)`
Advertisements
Solution
`tan^-1 x/y - tan^-1 (x - y)/(x + y) = tan^-1 [(x/y - (x - y)/(x + y))/(1 + (x/y)((x - y)/(x + y)))]`
= `tan^-1 [((x(x + y) - y(x - y))/(y(x + y)))/(1 + (x(x - y))/(y(x + y)))]`
= `tan^-1 [(((x^2 + xy - xy + y^2))/(y(x + y)))/((y(x + y) + x(x - y))/(y(x + y)))]`
= `tan^-1 [(x^2 + y^2)/(xy + y^2 + x^2 - xy)]`
= `tan^-1 [(x^2 + y^2)/(x^2 + y^2)]`
= `tan^-1(1) = pi/4`
APPEARS IN
RELATED QUESTIONS
If `tan^-1(2x)+tan^-1(3x)=pi/4`, then find the value of ‘x’.
Find the value of the following:
`tan 1/2 [sin^(-1) (2x)/(1+ x^2) + cos^(-1) (1-y^2)/(1+y^2)]`, |x| < 1, y > 0 and xy < 1
if `sin(sin^(-1) 1/5 + cos^(-1) x) = 1` then find the value of x
`cos^(-1) (cos (7pi)/6)` is equal to ______.
Prove `tan^(-1) 1/5 + tan^(-1) (1/7) + tan^(-1) 1/3 + tan^(-1) 1/8 = pi/4`
Prove that:
`cot^(-1) ((sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1- sinx))) = x/2, x in (0, pi/4)`
sin (tan–1 x), |x| < 1 is equal to ______.
Prove that `tan {pi/4 + 1/2 cos^(-1) a/b} + tan {pi/4 - 1/2 cos^(-1) a/b} = (2b)/a`
Solve for x : \[\cos \left( \tan^{- 1} x \right) = \sin \left( \cot^{- 1} \frac{3}{4} \right)\] .
Find the value of `tan(sin^-1 3/5 + cot^-1 3/2)`
Solve: `tan^-1x = cos^-1 (1 - "a"^2)/(1 + "a"^2) - cos^-1 (1 - "b"^2)/(1 + "b"^2), "a" > 0, "b" > 0`
Choose the correct alternative:
sin(tan–1x), |x| < 1 is equal to
Evaluate `cos[cos^-1 ((-sqrt(3))/2) + pi/6]`
Prove that `tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/((1 + x^2) - sqrt(1 - x^2))) = pi/2 + 1/2 cos^-1x^2`
If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression.
`tan[tan^-1("d"/(1 + "a"_1 "a"_2)) + tan^-1("d"/(21 + "a"_2 "a"_3)) + tan^-1("d"/(1 + "a"_3 "a"_4)) + ... + tan^-1("d"/(1 + "a"_("n" - 1) "a""n"))]`
If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals ______.
If `"cot"^-1 (sqrt"cos" alpha) - "tan"^-1 (sqrt"cos" alpha) = "x",` the sinx is equal to ____________.
`"cos"^-1["cos"(2"cot"^-1(sqrt2 - 1))]` = ____________.
`"sin"^-1 (1 - "x") - 2 "sin"^-1 "x" = pi/2`
