Home
Class 12
MATHS
cos[2cos^(-1).(1)/(5)+sin^(-1).(1)/(5)]=...

`cos[2cos^(-1).(1)/(5)+sin^(-1).(1)/(5)]`=

A

`(2sqrt(6))/(5)`

B

`-(2sqrt(6))/(5)`

C

`(1)/(5)`

D

`-(1)/(5)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the expression \( \cos\left(2 \cos^{-1}\left(\frac{1}{5}\right) + \sin^{-1}\left(\frac{1}{5}\right)\right) \), we can follow these steps: ### Step 1: Use the Inverse Trigonometric Identity We know that: \[ \sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2} \] For our case, we can express \( \cos^{-1}\left(\frac{1}{5}\right) \) in terms of \( \sin^{-1}\left(\frac{1}{5}\right) \): \[ \cos^{-1}\left(\frac{1}{5}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{5}\right) \] ### Step 2: Substitute into the Cosine Expression Now, substituting this into our expression: \[ \cos\left(2 \cos^{-1}\left(\frac{1}{5}\right) + \sin^{-1}\left(\frac{1}{5}\right)\right) = \cos\left(2\left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1}{5}\right)\right) + \sin^{-1}\left(\frac{1}{5}\right)\right) \] ### Step 3: Simplify the Argument This simplifies to: \[ \cos\left(\pi - 2 \sin^{-1}\left(\frac{1}{5}\right) + \sin^{-1}\left(\frac{1}{5}\right)\right) = \cos\left(\pi - \sin^{-1}\left(\frac{1}{5}\right)\right) \] ### Step 4: Use the Cosine Identity Using the identity \( \cos(\pi - x) = -\cos(x) \): \[ \cos\left(\pi - \sin^{-1}\left(\frac{1}{5}\right)\right) = -\cos\left(\sin^{-1}\left(\frac{1}{5}\right)\right) \] ### Step 5: Find Cosine using the Right Triangle Let \( \theta = \sin^{-1}\left(\frac{1}{5}\right) \). In a right triangle, if \( \sin(\theta) = \frac{1}{5} \), then the opposite side is 1 and the hypotenuse is 5. The adjacent side can be found using the Pythagorean theorem: \[ \text{Adjacent} = \sqrt{5^2 - 1^2} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6} \] Thus, \[ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2\sqrt{6}}{5} \] ### Step 6: Substitute Back Now substituting back into our expression: \[ -\cos\left(\sin^{-1}\left(\frac{1}{5}\right)\right) = -\frac{2\sqrt{6}}{5} \] ### Final Answer Thus, the final value is: \[ \cos\left(2 \cos^{-1}\left(\frac{1}{5}\right) + \sin^{-1}\left(\frac{1}{5}\right)\right) = -\frac{2\sqrt{6}}{5} \]
Promotional Banner

Topper's Solved these Questions

  • TRIGONOMETRIC FUNCTIONS

    TARGET PUBLICATION|Exercise Evaluation Test|34 Videos
  • TRIGONOMETRIC FUNCTIONS

    TARGET PUBLICATION|Exercise Critical Thinking|88 Videos
  • THREE DIMENSIONAL GEOMETRY

    TARGET PUBLICATION|Exercise Evaluation Test|9 Videos
  • VECTORS

    TARGET PUBLICATION|Exercise Evaluation Test|18 Videos

Similar Questions

Explore conceptually related problems

Prove that : cos^(-1).(3)/(5)+ cos^(-1).(12)/(13) = sin^(-1)((12)/(5))

Which of the following angles is greater ? theta_(1) = sin^(-1).(4)/(5) + sin^(-1).(1)/(3) and theta_(2) = cos^(-1).(4)/(5) + cos^(-1).(1)/(3)

If |z-25i|le15 , then |maximum arg(z) - minimum arg(z)| equals (A) (pi)/(2)+cos^(-1)((3)/(5)) (B) sin^(-1)((3)/(5))-cos^(-1)((3)/(5)) (C) 2cos^(-1)((4)/(5)) (D) 2cos^(-1)((1)/(5))

sin^(-1)(1/5)+cos^(-1)x=(pi)/(2)

If alpha=sin(sin^(-1)( 1/sqrt3)/(3)),beta=cos(cos^(-1)((1)/(sqrt(5)))-sin^(-1)((2)/(sqrt(5)))) then (beta^(2))/((3 alpha-4a^(3))^(2)) is

Evaluate: sin(cos^(-1)(3)/(5)+csc^(-1)(13)/(5))

sin[cos^(-1)(3/5)+tan^(-1)2]=

Evaluate :cos(2cos^(-1)x+sin^(-1)x) at x=(1)/(5)

sin(cos^(-1)(4/5))

cos^(-1)((63)/(65))+2tan^(-1)((1)/(5))=sin^(-1)(3/5)