Question

tan-¹(67/19)=tan-¹(5/13)+cos-¹(3/5)​

Answer 1

Answer:

L.H.S = R.H.S

Step-by-step explanation:

EXPLANATION IS IN THE PHOTO.

Answer 2

Answer:

Proof given below

Step-by-step explanation:

To prove:

tan⁻¹ $\frac{67}{19}$ = tan⁻¹ $\frac{5}{13}$ + cos⁻¹ $\frac{3}{5}$

Taking RHS,

tan⁻¹ $\frac{5}{13}$ + cos⁻¹ $\frac{3}{5}$

We know that cos⁻¹ $\frac{3}{5}$ = tan⁻¹ $\frac{4}{3}$ (using inverse trigonometric value)

=> tan⁻¹ $\frac{5}{13}$ + cos⁻¹ $\frac{3}{5}$ = tan⁻¹ $\frac{5}{13}$ + tan⁻¹ $\frac{4}{3}$

We know that tan⁻¹ x + tan⁻¹ y = tan⁻¹ $\big(\frac{x+y}{1-xy }\big)$

=> tan⁻¹ $\frac{5}{13}$ + tan⁻¹ $\frac{4}{3}$   = tan⁻¹ $\big(\frac{\frac{5}{15} + \frac{4}{3}}{1-\frac{5}{15}* \frac{4}{3} }\big)$

                                = tan⁻¹ $\big(\frac{\frac{15+52}{39}}{\frac{39-20}{39}}\big)$

                                = tan⁻¹ $\big(\frac{\frac{67}{39}}{\frac{19}{39}}\big)$

                               = tan⁻¹ $\big(\frac{67*39}{39*19}}\big)$

                               = tan⁻¹ $\big(\frac{67}{39}}\big)$

                               = LHS

As LHS = RHS,

Hence proved.