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प्रश्न
Find the value of `tan(sin^-1 3/5 + cot^-1 3/2)`
उत्तर
Let α = `sin^-1 (3/5)`
⇒ sin α = `3/5`
1 – sin2α = `1 - 9/25 = 16/25`
cos2α = `16/25`
⇒ cos α = `4/5`
tan α = `(3/5)/(4/5) = 3/4`
α = `tan^-1 (3/4)`
⇒ `sin^-1 (3/5) = tan^-1(3/4)`
⇒ `tan[sin^-1 (3/5) + cot^-1 (3/2)]`
= `tan[tan^-1 3/4 + tan^-1 2/3]`
= `tan[tan^-1 ((3/4 + 2/3)/(1 - 3/4 xx 2/3))]`
= `tan [tan^-1 (17/12)/(6/12)]`
= `tan[tan^-1 (17/6)]`
= `17/6`
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