Guys pls solve quez 9 pls
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Lunaticboyrk:
Is quez difficult for u ????
ya.... it's tough...
im doing it bt nt getting the answer...
Ok thnxx for trying u can do your work don't waste your time
by wen do u need the answer?
Before exam
Nxt mon
tmrw?
or the day after?
Okay
Answers
Answered by
3
I hope it will help you
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Phenomenal maan gye aapko!!!!
Thanks
Waise i have to thnx you naa........so thnx from me
kk
Answered by
3
There is a simpler solution than involving a lot of trigonometry. Instead of theta I use the symbol A for the angle.
CosecA - Sin A = a³ => 1 - sin²A = a³ Sin A (∵ cosecA =1/sinA
=> Cos²A = a³ Sin A --(1)
=> SinA = cos²A/a³ ---(2)
SecA - CosA = b³ => 1 - Cos²A = b³ CosA
=> Sin²A = b³ CosA -- (3)
=> cosA = Sin²A/b³ --(4)
Substitute the value of SinA from (1) in (3):
=> (cos²A/a³)² = b³ CosA
=> Cos³A = b³ a⁶
=> CosA = b a² --- (5)
Substitute the value of CosA from (4) in (1) :
=> Sin³A = a³ b⁶
=> SinA = a b² ---(6)
Square the terms in (5) and (6)
=> (ba²)² + (ab²)² = 1
=> a² b² (a² + b²) = 1
CosecA - Sin A = a³ => 1 - sin²A = a³ Sin A (∵ cosecA =1/sinA
=> Cos²A = a³ Sin A --(1)
=> SinA = cos²A/a³ ---(2)
SecA - CosA = b³ => 1 - Cos²A = b³ CosA
=> Sin²A = b³ CosA -- (3)
=> cosA = Sin²A/b³ --(4)
Substitute the value of SinA from (1) in (3):
=> (cos²A/a³)² = b³ CosA
=> Cos³A = b³ a⁶
=> CosA = b a² --- (5)
Substitute the value of CosA from (4) in (1) :
=> Sin³A = a³ b⁶
=> SinA = a b² ---(6)
Square the terms in (5) and (6)
=> (ba²)² + (ab²)² = 1
=> a² b² (a² + b²) = 1
clik on thanks ... select best an
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