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प्रश्न
If sin A = `(7)/(25)`, find the value of : cot2A - cosec2A
उत्तर
Consider ΔABC, where ∠B = 90°
⇒ sin A = `"Perpendicular"/"Hypotenuse" = "BC"/"AC" = (7)/(25)`
⇒ cosec A = `(1)/"sin A" = (25)/(7)`
By Pythagoras theorem,
AC2 = AB2 + BC2
⇒ AB2
= AC2 - BC2
= 252 - 72
= 625 - 49
= 576
⇒ AB - 24
Now,
cos A = `"Base"/"Hypotenuse" = "AB"/"AC" = (24)/(25)`
tan A = `"Perpendicular"/"Base" = "BC"/"AB" = (7)/(24)`
⇒ cot A = `(1)/"tan A" = (24)/(7)`
cot2A - cosec2A
= `(24/7)^2 - (25/7)^2`
= `(576)/(49) - (625)/(49)`
= `(-49)/(49)`
= -1.
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