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प्रश्न
If secA = `(5)/(4)`, cerify that `(3sin"A" - 4sin^3"A")/(4cos^3"A" - 3cos"A") = (3tan"A" - tan^3"A")/(1 - 3tan^2"A")`.
उत्तर
sec A = `(5)/(4)`
⇒ cos A = `(4)/(5) = "Base"/"Hypotenuse"`
Perpendicular
= `sqrt(("Hypotenuse")^2 - ("Base")^2`
= `sqrt(25 - 16)`
= `sqrt(9)`
= 3
sin A = `"Perpendicular"/"Hypotenuse" = (3)/(5)`
tan A = `"Perpendicular"/"Base" = (3)/(4)`
To show: `(3sin"A" - 4sin^3"A")/(4cos^3"A" - 3cos"A") = (3tan"A" - tan^3"A")/(1 - 3tan^2"A")`.
L.H.S. = `(3sin"A" - 4sin^3"A")/(4cos^3"A" - 3cos"A")`
= `(3(3/5) - 4(3/5)^3)/(4(4/5)^3 - 3(4/5)`
= `(9/5 - 108/125)/(256/125 - 12/5)`
= `((225 - 108)/125)/((256 - 300)/125)`
= `(117)/(-44)`
R.H.S. = `(3tan"A" - tan^3"A")/(1 - 3tan^2"A")`
= `(3(3/4) - (3/4)^3)/(1 - 3(3/4)^2`
= `(9/4 - 27/64)/(1 - 27/16)`
= `((144 - 27)/64)/((16 - 27)/16)`
= `(117)/(64) xx (16)/(-11)`
= `(117)/(4) xx (1)/(-11)`
= `(-117)/(44)`
⇒ L.H.S. = R.H.S.
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