Question

Show that tan^-1 (1÷2) + tan^-1 (2÷11) = tan^-1 (3÷4​)

Answer 1

Step-by-step explanation:

LHS = RHS

HENCE PROVED

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Answer 2

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

$tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{2}{11}=tan^{-1} \dfrac{3}{4}$

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• LHS = RHS

$\Large{\underline{\underline{\bf{Formula\:used:}}}}$

$tan^{-1}x+tan^{-1}y=tan^{-1}\dfrac{x+y}{1-xy}$

$\Large{\underline{\underline{\bf{Proof:}}}}$

$tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{2}{11}=tan^{-1} \dfrac{\dfrac{1}{2} +\dfrac{2}{11} }{1-\dfrac{1}{2}\times \dfrac{2}{11} }$

$tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{2}{11}=tan^{-1} \frac{\dfrac{11+4}{22} }{1-\dfrac{2}{22} }$

$tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{2}{11}=tan^{-1} \dfrac{\dfrac{15}{22} }{\dfrac{20}{22} }$

$tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{2}{11}=tan^{-1} \dfrac{15}{20}$

$tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{2}{11}=tan^{-1} \dfrac{3}{4}$

= RHS

Hence proved.