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प्रश्न
Prove the following:
`tan^-1(1/2) + tan^-1(1/3) = pi/(4)`
योग
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उत्तर
L.H.S. = `tan^-1(1/2) + tan^-1(1/3)`
= `tan^-1 [(1/2 + 1/3)/(1 - 1/2 * 1/3)]` ...since `1/2 > 0, 1/3 > 0` and `(1/2)(1/3) < 1`
= `tan ^-1 ((5/6)/(1 - 1/6))`
= `tan^-1((5/6)/(5/6))`
= tan-1(1)
= `tan^-1(tan pi/4)`
= `pi/(4)`
= R.H.S.
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