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प्रश्न
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
उत्तर
LHS=`tan(cos^-1 4/5+tan^-1 2/3)=tan(tan^-1 sqrt(1-(4/5)^2)/(4/5)+tan^-1 2/3)` `[thereforecos^-1x=tan^-1(sqrt(1-x^2)/x)]`
`=tan(tan^-1 3/4+tan^-1 2/3)`
`=tan[tan^-1((3/4+2/3)/(1-3/4xx2/3))]` `[thereforetan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))]`
`=tan[tan^-1((17/12)/(6/12))`
`=tan[tan^-1 17/6]`
`=17/6=`RHS
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