Advertisements
Advertisements
प्रश्न
Find the value of `cot(tan^(-1) a + cot^(-1) a)`
उत्तर
`cot(tan^(-1) a + cos^(-1))`
`= cot(pi/2) [tan^(-1) x + cot^(-1) x = pi/2]`
= 0
APPEARS IN
संबंधित प्रश्न
Prove that `cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))=x/2;x in (0,pi/4) `
Prove that:
`tan^(-1)""1/5+tan^(-1)""1/7+tan^(-1)""1/3+tan^(-1)""1/8=pi/4`
If a line makes angles 90°, 60° and θ with x, y and z-axis respectively, where θ is acute, then find θ.
Prove `2 tan^(-1) 1/2 + tan^(-1) 1/7 = tan^(-1) 31/17`
Find the value of 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 `tan^(-1) (x-1)/(x - 2) + tan^(-1) (x + 1)/(x + 2) = pi/4` then find the value of x.
Find the value of the given expression.
`tan^(-1) (tan (3pi)/4)`
Solve the following equation:
`2 tan^(-1) (cos x) = tan^(-1) (2 cosec x)`
Prove that `tan {pi/4 + 1/2 cos^(-1) a/b} + tan {pi/4 - 1/2 cos^(-1) a/b} = (2b)/a`
Prove that
\[2 \tan^{- 1} \left( \frac{1}{5} \right) + \sec^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) = \frac{\pi}{4}\] .
Find the value, if it exists. If not, give the reason for non-existence
`tan^-1(sin(- (5pi)/2))`
Find the value of `tan(sin^-1 3/5 + cot^-1 3/2)`
Prove that `sin^-1 3/5 - cos^-1 12/13 = sin^-1 16/65`
If tan–1x + tan–1y + tan–1z = π, show that x + y + z = xyz
Solve: `sin^-1 5/x + sin^-1 12/x = pi/2`
Choose the correct alternative:
If `sin^-1x + sin^-1y = (2pi)/3` ; then `cos^-1x + cos^-1y` is equal to
Choose the correct alternative:
sin–1(2 cos2x – 1) + cos–1(1 – 2 sin2x) =
Choose the correct alternative:
sin(tan–1x), |x| < 1 is equal to
Evaluate `tan^-1(sin((-pi)/2))`.
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 cos–1x > sin–1x, then ______.
If y = `2 tan^-1x + sin^-1 ((2x)/(1 + x^2))` for all x, then ______ < y < ______.
If `"tan"^-1 ("cot" theta) = 2theta, "then" theta` is equal to ____________.
`"tan"^-1 1 + "cos"^-1 ((-1)/2) + "sin"^-1 ((-1)/2)`
`"sin" {2 "cos"^-1 ((-3)/5)}` is equal to ____________.
`"cos" (2 "tan"^-1 1/7) - "sin" (4 "sin"^-1 1/3) =` ____________.
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:
`"cos" (2 "tan"^-1 1/7) - "sin" (4 "sin"^-1 1/3) =` ____________.
The value of `"cos"^-1 ("cos" ((33pi)/5))` is ____________.
`"sin"^-1 (1/sqrt2)`
If `"sin" {"sin"^-1 (1/2) + "cos"^-1 "x"} = 1`, then the value of x is ____________.
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 ______.
`50tan(3tan^-1(1/2) + 2cos^-1(1/sqrt(5))) + 4sqrt(2) tan(1/2tan^-1(2sqrt(2)))` is equal to ______.
`tan(2tan^-1 1/5 + sec^-1 sqrt(5)/2 + 2tan^-1 1/8)` is equal to ______.
Solve:
sin–1(x) + sin–1(1 – x) = cos–1x.