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प्रश्न
Prove that:
`2sin^-1 3/5=tan^-1 24/7`
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उत्तर
`2sin^-1 3/5`
= `2tan^-1 3/sqrt(5^2 - 3^2) ...[sin^-1 p/h = tan^-1 p/sqrt(h^2 - p^2)]`
= `2 tan^-1 3/sqrt(25 - 9)`
= `2 tan^-1 3/sqrt16`
= `2tan^-1 3/4`
= `tan^-1 (2 xx 3/4)/(1 - (3/4)^2) ...[2tan^-1 x = tan^-1 (2x)/(1 - x^2)]`
= `tan^-1 (3/2)/(1 - 9/16)`
= `tan^-1 (3/2)/((16 - 9)/16)`
= `tan^-1 (3/2)/(7/16)`
= `tan^-1 (3/2 xx 16/7)`
= `tan^-1 (3/1 xx 8/7)`
= `tan^-1 24/7`
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