Advertisements
Advertisements
Question
Solve for x : \[\cos \left( \tan^{- 1} x \right) = \sin \left( \cot^{- 1} \frac{3}{4} \right)\] .
Solution
Given:
\[\cos \left( \tan^{- 1} x \right) = \sin \left( \cot^{- 1} \frac{3}{4} \right)\] .........(1)
\[cos\theta = \sin\left( \frac{\pi}{2} - \theta \right)\]
\[ \Rightarrow \cos\left( \tan^{- 1} x \right) = \sin\left( \frac{\pi}{2} - \tan^{- 1} x \right)\]
\[ \Rightarrow \cos\left( \tan^{- 1} x \right) = \sin\left( \cot^{- 1} x \right)\]
Substituting the value of
\[\sin\left( \cot^{- 1} x \right) = \sin\left( \cot^{- 1} \frac{3}{4} \right)\]
\[ \Rightarrow x = \frac{3}{4}\]
APPEARS IN
RELATED QUESTIONS
Prove `2 tan^(-1) 1/2 + tan^(-1) 1/7 = tan^(-1) 31/17`
Write the function in the simplest form: `tan^(-1) 1/(sqrt(x^2 - 1)), |x| > 1`
`cos^(-1) (cos (7pi)/6)` is equal to ______.
Prove that:
`cos^(-1) 12/13 + sin^(-1) 3/5 = sin^(-1) 56/65`
Prove that
\[2 \tan^{- 1} \left( \frac{1}{5} \right) + \sec^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) = \frac{\pi}{4}\] .
Prove that `sin^-1 3/5 - cos^-1 12/13 = sin^-1 16/65`
Prove that `tan^-1x + tan^-1y + tan^-1z = tan^-1[(x + y + z - xyz)/(1 - xy - yz - zx)]`
Prove that `tan^-1x + tan^-1 (2x)/(1 - x^2) = tan^-1 (3x - x^3)/(1 - 3x^2), |x| < 1/sqrt(3)`
If |x| ≤ 1, then `2 tan^-1x + sin^-1 ((2x)/(1 + x^2))` is equal to ______.
If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals ______.
If cos–1x > sin–1x, then ______.
Solve for x : `"sin"^-1 2 "x" + sin^-1 3"x" = pi/3`
If `"tan"^-1 ("cot" theta) = 2theta, "then" theta` is equal to ____________.
The value of cot-1 9 + cosec-1 `(sqrt41/4)` is given by ____________.
If x = a sec θ, y = b tan θ, then `("d"^2"y")/("dx"^2)` at θ = `π/6` is:
The value of `"tan"^-1 (1/2) + "tan"^-1(1/3) + "tan"^-1(7/8)` is ____________.
`"sin"^-1 (1 - "x") - 2 "sin"^-1 "x" = pi/2`
`"cos"^-1 1/2 + 2 "sin"^-1 1/2` is equal to ____________.
`"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:
Measure of ∠DAB = ________.
Find the value of `cos^-1 (1/2) + 2sin^-1 (1/2) ->`:-
`50tan(3tan^-1(1/2) + 2cos^-1(1/sqrt(5))) + 4sqrt(2) tan(1/2tan^-1(2sqrt(2)))` is equal to ______.
The set of all values of k for which (tan–1 x)3 + (cot–1 x)3 = kπ3, x ∈ R, is the internal ______.
If `tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)` = then prove that 24x2 – 23x – 12 = 0
The value of cosec `[sin^-1((-1)/2)] - sec[cos^-1((-1)/2)]` is equal to ______.
Solve:
sin–1(x) + sin–1(1 – x) = cos–1x.
