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see attachment and explain it

 see attachment and explain it

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Grade:12

1 Answers

Latika Leekha
askIITians Faculty 165 Points
8 years ago
Given that cosec θ – sin θ
= b3
and sec θ – cos θ = a3
Solving the first equation we get,
1/ sin θ – sin θ = b3
This gives (1-sin2θ) / sin θ = b3
This yields cos2θ/ sin θ = b3
Similarly, solving the second equation on similar lines gives
sin2θ/ cos θ = a3
Now, considering L.H.S
a4b2 + a2b4
= (sin2θ/ cos θ)4/3 . (cos2θ/ sin θ)2/3 + (sin2θ/ cos θ)2/3 . (cos2θ/ sin θ)4/3
On simplifying this we get,
= sin 2θ + cos2θ
= 1.
Hence, a4b2 + a2b4 = 1.

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