Prove \(2\tan^{-1}\frac {1}{2}+\tan^{-1}\frac{1}{7}=\tan^{-1}\frac{31}{17}\)
To prove \(2\tan^{-1}\frac {1}{2}+\tan^{-1}\frac{1}{7}=\tan^{-1}\frac{31}{17}\)
LHS= \(2\tan^{-1}\frac {1}{2}+\tan^{-1}\frac{1}{7}\)
= \(\tan^{-1} \frac{\frac{2.1}{2}}{1-(\frac{1}{2})^2}+\tan^{-1}\frac{1}{7} \ [2tan^{-1x}=\tan^{-1}\frac{2x}{1-x^2}]\)
\(\tan^{-1}\frac{1}{(\frac{3}{4})}+\tan^{-1}\frac{1}{7}\)
\(\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{7}\)
= \(\tan^{-1}\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}.\frac{1}{7}}\)
=\(\tan^{-1}\frac{\frac{(28+3)}{21}}{\frac{(21-4)}{21}}\)
= \(\tan^{-1}\frac{31}{17}\)=RHS
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)