Advertisements
Advertisements
प्रश्न
Prove that `2tan^(-1)(1/5)+sec^(-1)((5sqrt2)/7)+2tan^(-1)(1/8)=pi/4`
Advertisements
उत्तर
`2tan^(-1)(1/5)+sec^(-1)((5sqrt2)/7)+2tan^(-1)(1/8)`
`=2tan^(-1)(1/5)+tan^(-1)(sqrt(((5sqrt2)/7)^2-1))+2tan^(-1)(1/8) [`
`=2tan^(-1)(1/5)+tan^(-1)(1/7)+2tan^(-1)(1/8)`
`=2(tan^(-1)(1/5)+tan^(-1)(1/8))+tan^(-1)(1/7)`
`=tan^(-1)((1/5+1/8)/(1-(1/5)xx(1/8)))+tan^(-1)(1/7) [`
`=2 tan^(−1)(13/39)+tan^(−1)(1/7)`
`=2 tan^(−1)(1/3)+tan^(−1)(1/7)`
`= tan^(-1)((2/3)/(1-1/9))+tan^(−1)(1/7) [`
`=tan^(-1)(3/4)+tan^(-1)(1/7)`
`=tan^(-1)((3/4+1/7)/(1-(3/4)xx(1/7)))`
`=tan^(-1)(1)`
`=pi/4`
`=RHS`
Hence proved.
संबंधित प्रश्न
Prove that: `tan^(-1)(1/2)+tan^(-1)(1/5)+tan^(-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
Write the function in the simplest form: `tan^(-1) ((cos x - sin x)/(cos x + sin x)) `,` 0 < x < pi`
`sin[pi/3 - sin^(-1) (-1/2)]` is equal to ______.
Prove that:
`cos^(-1) 12/13 + sin^(-1) 3/5 = sin^(-1) 56/65`
Prove `tan^(-1) 1/5 + tan^(-1) (1/7) + tan^(-1) 1/3 + tan^(-1) 1/8 = pi/4`
Solve for x : \[\cos \left( \tan^{- 1} x \right) = \sin \left( \cot^{- 1} \frac{3}{4} \right)\] .
Solve for x : `tan^-1 ((2-"x")/(2+"x")) = (1)/(2)tan^-1 ("x")/(2), "x">0.`
Find the value, if it exists. If not, give the reason for non-existence
`sin^-1 (cos pi)`
Find the value, if it exists. If not, give the reason for non-existence
`tan^-1(sin(- (5pi)/2))`
Find the value of the expression in terms of x, with the help of a reference triangle
sin (cos–1(1 – x))
Find the value of the expression in terms of x, with the help of a reference triangle
`tan(sin^-1(x + 1/2))`
Prove that `tan^-1x + tan^-1 (2x)/(1 - x^2) = tan^-1 (3x - x^3)/(1 - 3x^2), |x| < 1/sqrt(3)`
Evaluate `tan^-1(sin((-pi)/2))`.
Solve the equation `sin^-1 6x + sin^-1 6sqrt(3)x = - pi/2`
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 `sin^-1 ((2"a")/(1 + "a"^2)) + cos^-1 ((1 - "a"^2)/(1 + "a"^2)) = tan^-1 ((2x)/(1 - x^2))`. where a, x ∈ ] 0, 1, then the value of x is ______.
The number of real solutions of the equation `sqrt(1 + cos 2x) = sqrt(2) cos^-1 (cos x)` in `[pi/2, pi]` is ______.
The value of cot–1(–x) for all x ∈ R in terms of cot–1x is ______.
`"tan"^-1 1 + "cos"^-1 ((-1)/2) + "sin"^-1 ((-1)/2)`
`"tan"^-1 1/3 + "tan"^-1 1/5 + "tan"^-1 1/7 + "tan"^-1 1/8 =` ____________.
`"cos" (2 "tan"^-1 1/7) - "sin" (4 "sin"^-1 1/3) =` ____________.
`"sin"^-1 (1/sqrt2)`
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:
Domain and Range of tan-1 x = ________.
Find the value of `cos^-1 (1/2) + 2sin^-1 (1/2) ->`:-
The value of `tan^-1 (x/y) - tan^-1 (x - y)/(x + y)` is equal to
If `tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)` = then prove that 24x2 – 23x – 12 = 0
Solve for x: `sin^-1(x/2) + cos^-1x = π/6`
If sin–1x + sin–1y + sin–1z = π, show that `x^2 - y^2 - z^2 + 2yzsqrt(1 - x^2) = 0`
