Question

cot^-1[2tan(cos^-1(8/17))]+tan^-1[2tan(sin^-1(8/17))]=tan^-1(300/161)​

Answer 1

Given: cot^-1[2tan(cos^-1(8/17))]+tan^-1[2tan(sin^-1(8/17))]=tan^-1(300/161)​

To find:  Prove the above trigonometric terms.

Solution:

• Now lets consider the LHS side:

             cot^-1[2tan(cos^-1(8/17))]+tan^-1[2tan(sin^-1(8/17))]

• So, now assume

             A = cos^-1(8/17)

             cos A = 8/17

             B = sin^-1(8/17)

             sin B = 8/17

• Now, cos A = 8/17 => sin A = √1 - cos²A

          sin A = √1 - 64/289

                   = √225/289

                   = 15/17

• Similarly, sin B = 8/17 => cos B = √1 - sin²B

           cos B = √1 - 64/289

                     = √225/289

                     = 15/17

• So, tan A = 15/17/8/17 = 15/8 and tan B = 8/17/15/17 = 8/15

• So now, put these in the LHS we get:

                cot^-1[2tan(tan^-1 (15/8))] + tan^-1[2tan(tan^-1 (8/15))]

                cot^-1 [ 2 x 15/8 ] + tan^-1 [ 2 x 8/15 ]

                cot^-1(30/8) + tan^-1(16/15)

                tan^-1(4/15) + tan^-1(16/15)

                tan^-1((4/15 + 16/15) /( 1 - 4/15 x 16/15))

                tan^-1 ( 300/161) ...........................RHS

Answer:

         cot^-1[2tan(cos^-1(8/17))]+tan^-1[2tan(sin^-1(8/17))]=tan^-1(300/161)​

          Hence proved