Advertisements
Advertisements
प्रश्न
In ΔABC, prove that `("b"^2 - "c"^2)/"a" cos"A" + ("c"^2 - "a"^2)/"b" cos"B" + ("a"^2 - "b"^2)/"c" cos "C"` = 0
Advertisements
उत्तर
In ΔABC by cosine rule, we have
cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`, cosB = `("a"^2 + "c"^2 - "b"^2)/(2"ac")`, cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")`
∴ L.H.S. = `("b"^2 - "c"^2)/"a" cos"A" + ("c"^2 - "a"^2)/"b" cos"B" + ("a"^2 - "b"^2)/"c" cos "C"`
= `("b"^2 - "c"^2)/"a" xx ("b"^2 + "c"^2 - "a"^2)/(2"bc") + ("c"^2 - "a"^2)/"b" xx ("a"^2 + "c"^2 - "b"^2)/(2"ac") + ("a"^2 - "b"^2)/"c" xx ("a"^2 + "b"^2 - "c"^2)/(2"ab")`
= `1/(2"abc") ("b"^4 + "b"^2"c"^2 - "a"^2"b"^2 - "b"^2"c"^2 - "c"^4 + "a"^2"c"^2 + "a"^2"c"^2 + "c"^4 - "b"^2"c"^2 - "a"^4 - "a"^2"c"^2 + "a"^2"b"^2 + "a"^4 + "a"^2"b"^2 - "b"^4 + "b"^2"c"^2)`
= 0
= R.H.S
संबंधित प्रश्न
In Δ ABC with the usual notations prove that `(a-b)^2 cos^2(C/2)+(a+b)^2sin^2(C/2)=c^2`
In any ΔABC, with usual notations, prove that b2 = c2 + a2 – 2ca cos B.
With usual notations, in ΔABC, prove that a(b cos C − c cos B) = b2 − c2
The principal solutions of cot x = -`sqrt3` are .................
In , ΔABC prove that
`"sin"(("B" - "C")/2) = (("b" - "c")/"a") "cos"("A"/2)`
Find the Cartesian coordinates of the point whose polar coordinates are :
`(4, pi/2)`
Find the polar co-ordinates of the point whose Cartesian co-ordinates are.
`(sqrt(2), sqrt(2))`
In ΔABC, if cot A, cot B, cot C are in A.P. then show that a2, b2, c2 are also in A.P.
In any Δ ABC, prove the following:
a2 sin (B - C) = (b2 - c2) sin A.
In any Δ ABC, prove the following:
`("b" - "c")/"a" = (tan "B"/2 - tan "C"/2)/(tan "B"/2 +tan "C"/2)`
In Δ ABC, if a, b, c are in A.P., then show that cot `"A"/2, cot "B"/2, cot "C"/2` are also in A.P.
In Δ ABC, prove that a2 (cos2 B - cos2 C) + b2 (cos2 C - cos2 A) + c2 (cos2 A - cos2 B) = 0.
Show that `2 sin^-1 (3/5) = tan^-1(24/7)`
Show that
`tan^-1(1/5) + tan^-1(1/7) + tan^-1(1/3) + tan^-1 (1/8) = pi/4.`
Prove that `tan^-1 sqrt"x" = 1/2 cos^-1 ((1 - "x")/(1 + "x"))`, if x ∈ [0, 1]
If `tan^-1 (("x" - 1)/("x" - 2)) + tan^-1 (("x" + 1)/("x" + 2)) = pi/4`, find the value of x.
State whether the following equation has a solution or not?
cos 2θ = `1/3`
Solve: `tan^-1 ("1 - x"/"1 + x") = 1/2 (tan^-1 "x")`, for x > 0.
In ∆ABC, if cos A = `(sinB)/(2sinC)`, then ∆ABC is ______.
In ∆ABC, prove that ac cos B − bc cos A = a2 − b2
In ∆ABC, if sin2A + sin2B = sin2C, then show that a2 + b2 = c2
In ΔABC, a = 3, b = 4 and sin A = `3/4`, find ∠B
With usual notations, prove that `(cos "A")/"a" + (cos "B")/"b" + (cos "C")/"c" = ("a"^2 + "b"^2 + "c"^2)/(2"abc")`
In ∆ABC, if a = 13, b = 14, c = 15, then find the value of cos B
In ∆ABC, if `(2cos "A")/"a" + (cos "B")/"b" + (2cos"C")/"c" = "a"/"bc" + "b"/"ca"`, then show that the triangle is a right angled
In a triangle ABC with usual notations, if `(cos "A")/"a" = (cos "B")/"b" = (cos "C")/"c"`, then area of triangle ABC with a = `sqrt6` is ____________.
In a triangle ABC, If `(sin "A" - sin "C")/(cos "C" - cos "A")` = cot B, then A, B, C are in ________.
If one side of a triangle is double the other and the angles opposite to these sides differ by 60°, then the triangle is ______
If polar co-ordinates of a point are `(1/2, pi/2)`, then its cartesian co-ordinates are ______.
In ΔABC, if `"a" cos^2 "C"/2 + "c" cos^2 "A"/2 = (3"b")/2`, then a, b, c are in ______.
If a = 13, b = 14, c = 15, then `cos("A"/2)` = ______.
In a ΔABC, if `("b" + "c")/11 = ("c" + "a")/12 = ("a" + "b")/13`, then cos C = ______.
Let ABC be a triangle such that ∠A = 45°, ∠B = 75° then `"a" + "c"sqrt(2)` is equal to ______. (in usual notation)
In a triangle ABC, in usual notation, (a + b + c)(b + c – a) = λbc will be true if ______.
In a triangle ABC, ∠C = 90°, then `(a^2 - b^2)/(a^2 + b^2)` is ______.
In ΔABC, with usual notations, if a, b, c are in A.P. Then `a cos^2 (C/2) + c cos^2(A/2)` = ______.
The perimeter of ΔABC is 20, ∠A = 60°, area of ΔABC = `10sqrt(3)`, then find the values of a, b, c.
In ΔABC, a = 3, b = 1, cos(A – B) = `2/9`, find c.
In a triangle ABC, with usual notations, if \[\frac{2\cos\mathrm{A}}{\mathrm{a}}+\frac{\cos\mathrm{B}}{\mathrm{b}}+\frac{2\cos\mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ac}}\]then ∠A =
With usual notations, in a triangle ABC, if θ is any real number, then a cos(B - θ) + b cos (A + θ) is
