Advertisements
Advertisements
Question
If sin–1x + sin–1y + sin–1z = π, show that `x^2 - y^2 - z^2 + 2yzsqrt(1 - x^2) = 0`
Advertisements
Solution
Given, sin–1x + sin–1y + sin–1z = π
`\implies` sin–1x + sin–1y = π – sin–1z
`\implies sin^-1[xsqrt(1 - y^2) + ysqrt(1 - x^2)] = (pi - sin^-1z)`
`\implies xsqrt(1 - y^2) + ysqrt(1 - x^2) = sin(pi - sin^-1z)`
`\implies xsqrt(1 - y^2) + ysqrt(1 - x^2) = z`
`\implies xsqrt(1 - y^2) = z - ysqrt(1 - x^2)`
Now squaring on both sides, we get,
`(xsqrt(1 - y^2))^2 = (z - ysqrt(1 - x^2))^2`
`\implies x^2(1 - y^2) = (z^2 + y^2(1 - x^2) - 2zy sqrt(1 - x^2))`
`\implies x^2 - x^2y^2 = z^2 + y^2 - x^2y^2 - 2yz sqrt(1 - x^2)`
`\implies x^2 - y^2 - z^2 + 2yz sqrt(1 - x^2)` = 0
Hence proved
APPEARS IN
RELATED QUESTIONS
Prove that `2tan^(-1)(1/5)+sec^(-1)((5sqrt2)/7)+2tan^(-1)(1/8)=pi/4`
Write the following function in the simplest form:
`tan^(-1) (sqrt((1-cos x)/(1 + cos x)))`, 0 < x < π
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
Find the value of the given expression.
`tan(sin^(-1) 3/5 + cot^(-1) 3/2)`
Find the value of the expression in terms of x, with the help of a reference triangle
cos (tan–1 (3x – 1))
Prove that `tan^-1 2/11 + tan^-1 7/24 = tan^-1 1/2`
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^-1 (tan pi/4) - sin^-1 (sqrt(3/x)) = pi/6`. Then x is a root of the equation
Evaluate tan (tan–1(– 4)).
Evaluate: `tan^-1 sqrt(3) - sec^-1(-2)`.
Evaluate `cos[sin^-1 1/4 + sec^-1 4/3]`
Prove that cot–17 + cot–18 + cot–118 = cot–13
Solve the equation `sin^-1 6x + sin^-1 6sqrt(3)x = - pi/2`
If `tan^-1x = pi/10` for some x ∈ R, then the value of cot–1x is ______.
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 3 tan–1x + cot–1x = π, then x equals ______.
If y = `2 tan^-1x + sin^-1 ((2x)/(1 + x^2))` for all x, then ______ < y < ______.
Solve for x : `"sin"^-1 2 "x" + sin^-1 3"x" = pi/3`
If `"tan"^-1 ("cot" theta) = 2theta, "then" theta` is equal to ____________.
`"sin" {2 "cos"^-1 ((-3)/5)}` is equal to ____________.
The value of sin (2tan-1 (0.75)) is equal to ____________.
Simplest form of `tan^-1 ((sqrt(1 + cos "x") + sqrt(1 - cos "x"))/(sqrt(1 + cos "x") - sqrt(1 - cos "x")))`, `π < "x" < (3π)/2` is:
`"sin"^-1 ((-1)/2)`
The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the ground level. For viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following:
𝐴' Is another viewer standing on the same line of observation across the road. If the width of the road is 5 meters, then the difference between ∠CAB and ∠CA'B is ______.
The Simplest form of `cot^-1 (1/sqrt(x^2 - 1))`, |x| > 1 is
The value of cosec `[sin^-1((-1)/2)] - sec[cos^-1((-1)/2)]` is equal to ______.
Solve for x: `sin^-1(x/2) + cos^-1x = π/6`
