Prove that: `sin^(-1)(12)/(13)+cos^(-1)4/5+tan^(-1)(63)/(16)=pi`

Explanation

Let `sin^(-1)(12/13)=alpha,cos^(-1)(4/5)=beta,tan^(-1)(63/16)=gamma`
`rArrsinalpha=12/13,cosbeta=4/5,tangamma=63/16`
`rArr tanalpha=12/5`
`and tanbeta=3/4`
Now, `alpha=tan^(-1)(12/5)&beta=tan^(-1)(3/4)`
Here, `xy=12/5xx3/4=9/5 gt1`
`:.` We cannot use `tan^(-1)x +tan^(-1)y=tan^(-1)((x+y)/(1-xy)),xygt1`
Consider, `tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)`
`=(12/5+3/4)/(1-12/5xx3/4)`
`=(48+15)/(20-36)`
`=(-63)/16`
`=-tany`

`:' tanalpha =12/5&tanbeta=3/4`
`rArrtanalphatanbetagt1`
`(sinalphasinbeta)/(cosalphacosbeta)gt1`
`rArrsinalphasinbeta gt cosalphacosbeta ("as "cosalphacosbetagt0)`
`rArr cos(alpha- beta) lt0`
`pi/2ltalpha+betaltpi`
`:. tan(alpha+beta)=-tangamma`
`alpha+beta ne -gamma "as "alpha+beta in(pi/2,pi)`
`rArrtan(alpha+beta)=tan(pi-gamma)`
`rArralpha+beta=pi-gamma`
`rArralpha+gamma+beta=pi`
`rArr sin^(-1)(12/13)+cos^(-1)(4/5)+tan^(-1)(63/16)=pi`