Theorems and Laws [10]
In ΔABC, prove the following:
`(cos A)/a + (cos B)/b + (cos C)/c = (a^2 + b^2 + c^2)/(2abc)`
LHS = `(cos A)/a + (cos B)/b + (cos C)/c`
`= ((("b"^2 + "c"^2 - "a"^2)/"2bc"))/"a" + ((("c"^2 + "a"^2 - "b"^2)/"2ca"))/"b" + ((("a"^2 + "b"^2 - "c"^2)/"2ab"))/"c"`
`= ("b"^2 + "c"^2 - "a"^2)/"2abc" + ("c"^2 + "a"^2 - "b"^2)/"2abc" + ("a"^2 + "b"^2 - "c"^2)/"2abc"`
`= ("b"^2 + "c"^2 - "a"^2 + "c"^2 + "a"^2 - "b"^2 + "a"^2 + "b"^2 - "c"^2)/"2abc"`
`= ("a"^2 + "b"^2 + "c"^2)/"2abc"`
= RHS
LHS = `(cos A)/a + (cos B)/b + (cos C)/c`
= `(b cos A + a cos B)/(ab) + (cos C)/c`
= `c/(ab) + (cos C)/c` ...(By projection rule)
= `c/(ab) + (a^2 + b^2 - c^2)/(2 abc)` ...(By cosine rule)
= `(2c^2 + a^2 + b^2 - c^2)/(2 abc)`
= `(a^2 + b^2 + c^2)/(2 abc)` = R.H.S.
Prove that:
`tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x))) = pi/4 - 1/2 cos^-1 x`, for `- 1/sqrt2 ≤ x ≤ 1`
[Hint: Put x = cos 2θ]
Put x = cos θ
∴ θ = cos–1 x
L.H.S. = `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`
= `tan^-1 ((sqrt(1 + cos θ) - sqrt(1 - cos θ))/(sqrt(1 + cos θ) + sqrt(1 - cos θ)))`
= `tan^-1 [(sqrt(2 cos^2(θ/2)) - sqrt(2 sin^2 (θ/2)))/(sqrt(2 cos^2 (θ/2)) + sqrt(2 sin^2 (θ/2)))]`
= `tan^-1 [(sqrt(2) cos (θ/2) - sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2) + sqrt(2) sin (θ/2))]`
= `tan^-1 [((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) - (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))/((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) + (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))]`
= `tan^-1 [(1 - tan(θ/2))/(1 + tan (θ/2))]`
= `tan^-1 [(tan pi/4 - tan (θ/2))/(1 + tan pi/4. tan (θ/2))] ....[∵ tan pi/4 =1]`
= `tan^-1 [tan (pi/4 - θ/2)]`
= `pi/4 - θ/2`
= `pi/4 - 1/2 cos^-1`x .....[∵ θ = cos–1 x]
∴ L.H.S. = R.H.S.
Prove that:
`sin^(-1) 8/17 + sin^(-1) 3/5 = tan^(-1) 77/36`
`sin^-1 8/17 + sin^-1 3/5`
= `tan^-1 8/sqrt(17^2 - 8^2) + tan^-1 3/sqrt(5^2 - 3^2) ...[sin^-1 p/h = tan^-1 p/sqrt(h^2 - p^2)]`
= `tan^-1 8/sqrt(289 - 64) + tan^-1 3/sqrt(25 - 9)`
= `tan^-1 8/sqrt225 + tan^-1 3/sqrt16`
= `tan^-1 8/15 + tan^-1 3/4`
= `tan^-1 ((8/15 + 3/4)/(1 - 8/15 xx 3/4)) ...[tan^-1x + tan^-1y = tan^-1((x + y)/(1 - x xx y))]`
= `tan^-1[((32 + 45)/60)/(1 - 24/60)]`
= `tan^-1 77/36`
Prove that:
`cos^(-1) 12/13 + sin^(-1) 3/5 = sin^(-1) 56/65`
Let x = `cos^(-1) 12/13` and y = `sin^(-1) 3/5`
or cos x = `12/13` and sin y = `3/5`
sin x = `sqrt (1 - cos^2 x)` and cos y = `sqrt(1 - sin^2 y)`
Now, sin x = `sqrt(1 - 144/169)` and cos y = `sqrt( 1 - 9/25)`
⇒ sin x = `5/13` and cos y = `4/5`
We know that,
sin (x + y) = sin x cos y + cos x sin y
= `5/13 xx 4/5 + 12/13 xx 3/5 `
= `20/65 + 36/65 `
= `56/65`
⇒ x + y = `sin ^-1(56/65)`
or, `cos^-1(12/13) + sin^-1 (3/5)`
= `sin^-1(56/65)`
Prove that:
`tan^(-1) 63/16 = sin^(-1) 5/13 + cos^(-1) 3/5`
Let `sin^(-1) 5/13` = x and `cos^(-1) 3/5` = y
⇒ sin x = `5/13 ` and cos y = `3/5`
or tan x = `5/12` and tan y = `4/3`
⇒ x = `tan^-1 5/12` and y = `tan^(-1) 4/3`
x + y = `tan^-1 5/12 + tan^-1 4/3`
= `tan^-1 ((5/12 + 4/3)/(1 - 5/12 xx 4/3))`
= `tan^(-1) ((15+48)/(36-20))`
= `tan^(-1) 63/16`
Prove that:
`cos^(-1) 4/5 + cos^(-1) 12/13 = cos^(-1) 33/65`
Let `cos^(-1) 4/5` = x
Then, cos x = `4/5`
⇒ sin x = `sqrt (1 - (4/5)^2)`
⇒ sin x = `sqrt(1 - 16/25)`
⇒ sin x = `sqrt(9/25)`
⇒ sin x = `3/5`
∴ tan x = `3/4` ⇒ x = `tan^(-1) 3/4`
∴ `cos^(-1) 4/5 = tan^(-1) 3/4` ...(1)
Now let `cos^(-1) 12/13` = y
Then cos y = `12/13`
⇒ sin y = `5/13`
∴ tan y = `5/12` ⇒ y = `tan^(-1) 5/12`
∴ `cos^(-1) 12/13 = tan^(-1) 5/12` ....(2)
Let `cos^(-1) 33/65` = z
Then cos z = `33/65`
⇒ sin z = `56/65`
∴ tan z = `56/33` ⇒ z = `tan^(-1) 56/33`
∴ `cos^(-1) 33/65 = tan^(-1) 56/33` ....(3)
Now, we will prove that:
L.H.S = `cos^(-1) 4/5 + cos^(-1) 12/13`
= `tan^(-1) 3/4 + tan^(-1) 5/12` ....[Using (1) and (2)]
= `tan^(-1) (3/4 + 5/12)/(1 - 3/4 * 5/12) ....[tan^(-1) x + tan^(-1) y = tan^(-1) (x + y)/(1 - xy)]`
= `tan^(-1) (36+20)/(48-15)`
= `tan^(-1) 56/33`
= `cos^(-1) 33/65` .....[by (3)]
= R.H.S.
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)`
L.H.S. = `cot^(-1) ((sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1- sinx)))`
= `cot^(-1) (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x)) xx (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x))`
= `cot^(-1) ((1+sinx) + (1-sinx) + 2sqrt(1 - sin^2 x))/((1+sinx) - (1 - sinx)`
= `cot^(-1) (2(1 + cos x))/(2sin x)`
= `cot^(-1) (1+ cosx)/sin x`
= `cot^(-1) (2 cos^2 x/2)/(2sin x/2 cos x/2)`
= `cot^-1 (cot x/2)`
= `x/2`
L.H.S. = R.H.S.
Prove the following:
3 sin−1 x = sin−1 (3x − 4x3), `x ∈ [-1/2, 1/2]`
Let x = sin θ
Then, sin−1 x = θ
We have
R.H.S = sin−1 (3x − 4x3) = sin−1 (3 sin θ − 4 sin3θ)
= sin−1 (sin 3θ) = sin−1 (3 sin θ − 4 sin3θ)
= 3θ = sin−1 (3 sin θ − 4 sin3θ)
= 3 sin−1 x = sin−1 (3 sin θ − 4 sin3θ)
R.H.S = L.H.S
Prove that:
`tan^(-1) sqrtx = 1/2 cos^(-1) (1-x)/(1+x)`, x ∈ [0, 1]
Let x = tan2 θ
⇒ `sqrtx` = tan θ
⇒ θ = `tan^(-1) sqrtx` ...(1)
∴ `(1-x)/(1+x)`
= `(1-tan^2 θ)/(1 + tan^2 θ)`
= cos 2θ
Now we have,
R.H.S = `1/2 cos^(-1) (1-x)/(1+x)`
= `1/2 cos^(-1)(cos 2θ)`
= `1/2 xx 2θ`
= θ
= `tan^(-1) sqrtx` ....[From (1)]
R.H.S. = L.H.S.
Prove the following:
3cos−1x = cos−1(4x3 − 3x), `x ∈ [1/2, 1]`
Let x = cos θ
Then, cos−1x = θ
We have,
R.H.S = cos−1(4x3 − 3x)
⇒ cos−1(4 cos3θ − 3 cos θ)
⇒ cos−1(cos 3θ) = cos−1(4x3 − 3x)
⇒ 3θ = cos−1(4x3 − 3x)
⇒ 3 cos−1x = cos−1(4x3 − 3x)
R.H.S = L.H.S
