हिंदी

Prove the following : cos-1(35)+cos-1(45)=π2 - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Prove the following:

`cos^-1(3/5) + cos^-1(4/5) = pi/(2)`

योग
Advertisements

उत्तर

Let `cos^-1(3/5)` = x

∴ cos x = `(3)/(5), "where"  0 < x < pi/(2)`
∴ sin x > 0

Now,

sin x = `sqrt(1 - cos^2x)`

= `sqrt(1 - 9/25)`

= `sqrt(16/25)`

= `(4)/(5)`
∴ x = `sin^-1(4/5)`

∴ `cos^-1(3/5) = sin^-1(4/5)`    ...(1)

L.H.S. = `cos^-1(3/5) + cos^-1(4/5)`

= `sin^-1(4/5) + cos^-1(4/5)`   ...[By (1)]

= `pi/(2)                      ...[∵ sin^-1x + cos^-1x = pi/2]`

= R.H.S.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometric Functions - Exercise 3.3 [पृष्ठ १०३]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Exercise 3.3 | Q 3.4 | पृष्ठ १०३

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

Find the principal value of the following:

`cos^(-1) (sqrt3/2)`


Find the principal value of the following:

`cos^(-1) (-1/sqrt2)`


If sin−1 x = y, then ______.


Find the domain of the following function:

`f(x)=sin^-1x^2`

 


Evaluate the following:

`tan^-1(tan  (5pi)/6)+cos^-1{cos((13pi)/6)}`


Find the domain of `f(x)=cotx+cot^-1x`


Evaluate the following:

`cot^-1  1/sqrt3-\text(cosec)^-1(-2)+sec^-1(2/sqrt3)`


Evaluate: tan `[ 2 tan^-1  (1)/(2) – cot^-1 3]`


In ΔABC, if a = 18, b = 24, c = 30 then find the values of cos `A/2`


Find the principal value of the following: tan- 1( - √3)


Find the principal value of the following: sin-1 `(1/sqrt(2))`


Find the principal value of the following: cos- 1`(-1/2)`


Prove that sin `[tan^-1 ((1 - x^2)/(2x)) + cos^-1 ((1 - x^2)/(1 + x^2))]` = 1


Prove that cot−1(7) + 2 cot−1(3) = `pi/4`


Show that `tan^-1 (1/2) + tan^-1 (2/11) = tan^-1 (3/4)`


Solve: tan-1 (x + 1) + tan-1 (x – 1) = `tan^-1 (4/7)`


`sin^-1x + sin^-1  1/x + cos^-1x + cos^-1  1/x` = ______


The principle solutions of equation tan θ = -1 are ______ 


In Δ ABC, with the usual notations, if sin B sin C = `"bc"/"a"^2`, then the triangle is ______.


If `sin^-1x + cos^-1y = (3pi)/10,` then `cos^-1x + sin^-1y =` ______ 


The principal value of `tan^{-1(sqrt3)}` is ______  


`sin^2(sin^-1  1/2) + tan^2 (sec^-1  2) + cot^2(cosec^-1  4)` = ______.


If sin `(sin^-1  1/3 + cos^-1 x) = 1`, then the value of x is ______.


If `3sin^-1((2x)/(1 + x^2)) - 4cos^-1((1 - x^2)/(1 + x^2)) + 2tan^-1((2x)/(1 - x^2)) = pi/3`, then x is equal to ______ 


`sin{tan^-1((1 - x^2)/(2x)) + cos^-1((1 - x^2)/(1 + x^2))}` is equal to ______ 


Show that `cos(2tan^-1  1/7) = sin(4tan^-1  1/3)`


All trigonometric functions have inverse over their respective domains.


`"cos"  2 theta` is not equal to ____________.


`"sin"^2 25° +  "sin"^2 65°` is equal to ____________.


`("cos" 8° -  "sin" 8°)/("cos" 8° +  "sin" 8°)`  is equal to ____________.


`"cos"^-1 1/2 + 2  "sin"^-1  1/2` is equal to ____________.


If tan-1 (x – 1) + tan-1 x + tan-1 (x + 1) = tan-1 3x, then the values of x are ____________.


If 6sin-1 (x2 – 6x + 8.5) = `pi`, then x is equal to ____________.


`"sin" ["cot"^-1 {"cos" ("tan"^-1  "x")}] =` ____________.


`"cos"^-1 ["cos" (2  "cot"^-1 (sqrt2 - 1))] =` ____________.


`"tan"^-1 sqrt3 - "sec"^-1 (-2)` is equal to ____________.


If a = `(2sin theta)/(1 + costheta + sintheta)`, then `(1 + sintheta - costheta)/(1 + sintheta)` is 


what is the value of `cos^-1 (cos  (13pi)/6)`


Let x = sin–1(sin8) + cos–1(cos11) + tan–1(tan7), and x = k(π – 2.4) for an integer k, then the value of k is ______.


Number of values of x which lie in [0, 2π] and satisfy the equation

`(cos  x/4 - 2sinx) sinx + (1 + sin  x/4 - 2cosx)cosx` = 0


If ax + b (sec (tan–1 x)) = c and ay + b (sec.(tan–1 y)) = c, then `(x + y)/(1 - xy)` = ______.


If y = `tan^-1  (sqrt(1 + x^2) - sqrt(1 - x^2))/(sqrt(1 + x^2) + sqrt(1 - x^2))`, then `dy/dx` is equal to ______.


If sin–1x – cos–1x = `π/6`, then x = ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×