Prove \(\tan^{-1}\frac{2}{11}+\tan^{-1}\frac{7}{24}=\tan^{-1}\frac{1}{2}\)
To prove: \(\tan^{-1}\frac{2}{11}+\tan^{-1}\frac{7}{24}=\tan^{-1}\frac{1}{2}\)
LHS= \(\tan^{-1}\frac{2}{11}+\tan^{-1}\frac{7}{24}\)
= \(\tan^{-1}\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}+\frac{7}{24}}\) \(\bigg[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\bigg]\)
= \(\tan^{-1}\frac{48-77}{264-14}=\tan^{-1}\frac{125}{250}=\tan^{-1}\frac{1}{2}=R.H.S.\)
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)