Find the values of \(tan^{-1}(\tan\frac{3\pi}{4})\)
\(tan^{-1}(\tan\frac{3\pi}{4})\)
We know that \(\tan^{-1}(\tan x)=x \,if x\in(-\frac{\pi}{2},\frac{\pi}{2})\), which is the principal value branch of \(\tan^{-1}x\).
Here, 3π/4∉(-π/2'π/2)
Now,tan-1(tan(3π/4)can be written as:
tan-1(tan(3π/4)=tan-1[-tan(-3π/4)]=tan-1[-tan(π-π/4)]
=tan-1[-tanπ/4]=tan-1[tan(-π/4)] where,-π/4∈(-π/2'π/2)
therefore tan-1(tan(3π/4)=tan-1(tan(-π/4)=-π/4
The equation \[ 2 \cos^{-1} x = \sin^{-1} \left( 2 \sqrt{1 - x^2} \right) \] is valid for all values of \(x\) satisfying:
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)