मराठी

Evaluate: `Cos(Tan^-1 3/4)` - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate:

`cos(tan^-1  3/4)`

Advertisements

उत्तर

We have

`cos(tan^-1  3/4)=cos[1/2cos^-1((1-(3/4)^2)/(1+(3/4)^2))]`   `[therefore 2tan^-1x+cos^-1((1-x^2)/(1+x^2))]`

`=cos[1/2cos^-1(7/25)]`

Let

`y=cos^-1(7/25)`

`=>cosy=7/25`

Now,

`cos[1/2cos^-1(7/25)]=cos[1/2y]`

`=sqrt((cosy+1)/2)`    `[thereforecos2x=2cos^2x-1]`

`=sqrt((7/25+1)/2)`

`=sqrt(32/50)`

`=4/5`

`therefore cos[tan^-1(3/4)]=4/5`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.09 [पृष्ठ ५८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.09 | Q 2.3 | पृष्ठ ५८

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

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`


Solve the following for x:

`sin^(-1)(1-x)-2sin^-1 x=pi/2`


If (tan1x)2 + (cot−1x)2 = 5π2/8, then find x.


​Find the principal values of the following:

`cos^-1(sin   (4pi)/3)`


​Find the principal values of the following:

`cos^-1(tan  (3pi)/4)`


`sin^-1(sin3)`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`tan^-1(tan  (9pi)/4)`


Evaluate the following:

`sec^-1(sec  (5pi)/4)`


Evaluate the following:

`cosec^-1(cosec  (6pi)/5)`


Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`


Evaluate the following:

`cot^-1(cot  (9pi)/4)`


Evaluate the following:

`cot^-1{cot  ((21pi)/4)}`


Write the following in the simplest form:

`sin{2tan^-1sqrt((1-x)/(1+x))}`


Evaluate the following:

`sin(sec^-1  17/8)`


Prove the following result

`tan(cos^-1  4/5+tan^-1  2/3)=17/6`


Prove the following result:

`tan^-1  1/4+tan^-1  2/9=sin^-1  1/sqrt5`


Find the value of `tan^-1  (x/y)-tan^-1((x-y)/(x+y))`


Solve the following equation for x:

`tan^-1  (x-2)/(x-1)+tan^-1  (x+2)/(x+1)=pi/4`


If `cos^-1  x/2+cos^-1  y/3=alpha,` then prove that  `9x^2-12xy cosa+4y^2=36sin^2a.`


`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`


Solve the following equation for x:

`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`


If −1 < x < 0, then write the value of `sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))`


Write the value of

\[\cos^{- 1} \left( \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\].


Write the value of sin−1

\[\left( \sin( -{600}°) \right)\].

 

 


Write the value of sin \[\left\{ \frac{\pi}{3} - \sin^{- 1} \left( - \frac{1}{2} \right) \right\}\]


Write the value of tan1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]


Write the value of \[\sin^{- 1} \left( \frac{1}{3} \right) - \cos^{- 1} \left( - \frac{1}{3} \right)\]


Write the principal value of `sin^-1(-1/2)`


Write the principal value of \[\cos^{- 1} \left( \cos\frac{2\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{2\pi}{3} \right)\]


Write the value of \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]


Write the value of \[\sec^{- 1} \left( \frac{1}{2} \right)\]


Find the value of \[\tan^{- 1} \left( \tan\frac{9\pi}{8} \right)\]


sin\[\left[ \cot^{- 1} \left\{ \tan\left( \cos^{- 1} x \right) \right\} \right]\]  is equal to

 

 

It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\]   (−7), then the value of x is

 


The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]

 


Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .


Write the value of \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×