Question

Siddh kijiye ki tan-1 1/2 + tan-1 2/11= tan-1 3/4

Answer 1

Step-by-step explanation:

Given:

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

To find: Prove the equation

Solution:

Formula used:

$\boxed{\bold{\red{ {tan}^{ - 1} x + {tan}^{ - 1} y = {tan}^{ - 1} \left( \frac{x + y}{1 - xy} \right)}}}$

Apply the formula in LHS

${tan}^{ - 1} \frac{1}{2} + {tan}^{ - 1} \frac{2}{11} = {tan}^{ - 1} \left( \frac{ \frac{1}{2} + \frac{2}{11} }{1 - \frac{1}{2}. \frac{2}{11} } \right) \\ \\ = {tan}^{ - 1} \left( \frac{ \frac{11\times 1+2\times 2}{22} }{ \frac{22 - 2}{22} } \right) \\ \\ = {tan}^{ - 1} \left( \frac{15}{22} \times \frac{22}{20} \right) \\ \\ = {tan}^{ - 1} \left( \frac{15}{20} \right) \\ \\ = {tan}^{ - 1} \left( \frac{3}{4} \right) \\ \\ = RHS$

Final answer:

$\bold{\green{{tan}^{ - 1} \frac{1}{2} + {tan}^{ - 1} \frac{2}{11} = {tan}^{ - 1} \frac{3}{4}} }$

has been proved.

Hope it helps you.