If x?=np,x?=(2n+1)p/2?.n?Z, then Sin-1(Cosx)+Cos-1(Sinx)/Tan-1(Cotx)+Cot-1(Tanx)?
Question:

If xnπ,x(2n+1)π2.nZ,x\neq n \pi ,\, x \neq\,(2n+1)\frac {\pi}{2}.n\in Z, then Sin1(Cosx)+Cos1(Sinx)Tan1(Cotx)+Cot1(Tanx)\frac {Sin^{-1}(Cos x) + Cos^{-1}(Sin x)}{Tan ^{-1}(Cot x)+ Cot^{-1}(Tan x)} =

Updated On: Apr 2, 2024
  • π6\frac {\pi} {6}
  • π3\frac {\pi} {3}
  • π2\frac {\pi} {2}
  • π4\frac {\pi} {4}
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The Correct Option is D

Solution and Explanation

xnπ,x(2n+1)π2,nZx \neq n \pi, x \neq(2 n+1) \frac{\pi}{2}, n \in Z
Then, sin1(cosx)+cos1(sinx)tan1(cotx)+cot1(tanx)\frac{\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)}{\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x)}
=sin1{sin(π2x)}+cos1{cos(π2x)}tan1{tan(π2x)}+cot1{cot(π2x)}=\frac{\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-x\right)\right\}+\cos ^{-1}\left\{\cos \left(\frac{\pi}{2}-x\right)\right\}}{\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-x\right)\right\}+\cot ^{-1}\left\{\cot \left(\frac{\pi}{2}-x\right)\right\}}
=(π2x)+(π2x)(π2x)+(π2x)=(π2xπ2x)=1=\frac{\left(\frac{\pi}{2}-x\right)+\left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)+\left(\frac{\pi}{2}-x\right)}=\left(\frac{\pi-2 x}{\pi-2 x}\right)=1
But from the option we take x=π4x=\frac{\pi}{4}
=sin1(cosπ4)+cos1(sinπ4)tan1(cotπ4)+cot1(tanπ4)=\frac{\sin ^{-1}\left(\cos \frac{\pi}{4}\right)+\cos ^{-1}\left(\sin \frac{\pi}{4}\right)}{\tan ^{-1}\left(\cot \frac{\pi}{4}\right)+\cot ^{-1}\left(\tan \frac{\pi}{4}\right)}
=sin1(12)+cos1(12)tan1(1)+cot1(1)=\frac{\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)}{\tan ^{-1}(1)+\cot ^{-1}(1)}
{sin1x+cos1x=π2tan1x+cot1x=π2}\begin{Bmatrix}\sin^{-1}x +\cos^{-1}&x=\frac{\pi}{2}\\ \because&\\ \tan^{-1} x +cot^{-1}&x=\frac{\pi}{2}\end{Bmatrix}
=π/2π/2=1=\frac{\pi / 2}{\pi / 2}=1
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Concepts Used:

Inverse Trigonometric Functions

The inverse trigonometric functions are also called arcus functions or anti trigonometric functions. These are the inverse functions of the trigonometric functions with suitably restricted domains. Specifically, they are the inverse functions of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle’s trigonometric ratios. Inverse trigonometric functions are widely used in engineering, navigation, physics, and geometry.

Domain and Range Of Inverse Functions

Considering the domain and range of the inverse functions, following formulas are important to be noted:

  • sin(sin−1x) = x, if -1 ≤ x ≤ 1
  • cos(cos−1x) = x, if -1 ≤ x ≤ 1
  • tan(tan−1x) = x, if -∞ ≤ x ≤∞
  • cot(cot−1x) = x, if -∞≤ x ≤∞
  • sec(sec−1x) = x, if -∞ ≤ x ≤ -1 or 1 ≤ x ≤ ∞
  • cosec(cosec−1x) = x, if -∞ ≤ x ≤ -1 or 1 ≤ x ≤ ∞

Also, the following formulas are defined for inverse trigonometric functions.

  • sin−1(sin y) = y, if -π/2 ≤ y ≤ π/2
  • cos−1(cos y) =y, if 0 ≤ y ≤ π
  • tan−1(tan y) = y, if -π/2 <y< π/2
  • cot−1(cot y) = y if 0<y< π
  • sec−1(sec y) = y, if 0 ≤ y ≤ π, y ≠ π/2

cosec−1(cosec y) = y if -π/2 ≤ y ≤ π/2, y ≠ 0