Question

3 tan inverse 1 by 2 + root 3 minus tan inverse half equals to tan inverse 1 by 3 prove that​

Answer 1

Answer:

I explained above in pic,if any doubt msg me

Answer 2

we have to prove that, 3tan-¹[1/(2 + √3)] - tan-¹[1/2] = tan-¹[1/3]

or we have to prove that 3tan-¹[1/(2 + √3) = tan-¹[1/2] + tan-¹[1/3]

LHS = 3tan-¹[1/(2 + √3)]

= 3tan-¹[1/(2 + √3) × (2 - √3)/(2 - √3)]

= 3tan-¹[(2 - √3)/(2² - √3²)]

= 3tan-¹[(2 - √3)]

we know, tan(π/12) = (2 - √3)

so, tan-¹[(2 - √3)] = π/12

= 3tan-¹[(2 - √3) = 3 × π/12 = π/4

RHS = tan-¹[1/2] + tan-¹[1/3]

we know, tan-¹x + tan-¹y = tan-¹[(x + y)/(1 - xy)]

= tan-¹[(1/2 + 1/3)/(1 - 1/2 × 1/3)]

= tan-¹[5/5]

= tan-¹(1)

= π/4

here LHS = RHS

hence, 3tan-¹[1/(2 + √3)] = tan-¹[1/2] + tan-¹[1/3]

therefore, 3tan-¹[1/(2 + √3)] - tan-¹[1/2] = tan-¹[1/3]