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प्रश्न
Show that `2 sin^-1 (3/5) = tan^-1(24/7)`
उत्तर
Let `2 sin^-1 (3/5)` = x
Then sin x = `3/5`, where `0 < "x" < pi/2`
∴ cos x > 0
Now, cos x = `sqrt(1 - sin^2"x") = sqrt(1 - 9/25) = sqrt(16/25) = 4/5`
∴ `tan "x" = "sin x"/"cos x" = (3/5)/(4/5) = 3/4`
∴ x = `tan^-1(3/4)`
∴ `sin^-1 (3/5) = tan^-1(3/4)`
Now, LHS = `2sin^-1 (3/5) = 2tan^-1(3/4)`
`= tan^-1 (3/4) + tan^-1(3/4)`
= `tan^-1 [(3/4 + 3/4)/(1 - 3/4 xx 3/4)] = tan^-1 [(12 + 12)/(16 - 9)]`
`= tan^-1(24/7)` = RHS
Alternative Method:
LHS = `2sin^-1 (3/5) = 2tan^-1(3/4)`
`= tan^-1 [(2(3/4))/(1 - (3/4)^2)] .....[because 2 tan^-1 "x" = tan^-1 ("2x"/(1 - "x"^2))]`
`= tan^-1 [(3/2)/(1 - (9/16))]`
`= tan^-1 (3/2 xx 16/7)`
`= tan^-1 (24/7)`
= RHS
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