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vikash kumar

· started a discussion

· 1 Months ago

discussion releated (Question: 195903.0 )

discussion is not clear

Question:

If A cos2 \(\theta\)+ B sin2\(\theta\) =  \(\cfrac{sin^2 \ \theta \ (sec^2\theta \ + 1 )}{sec^2 \ \theta- 1}\)     then cot\(\theta\) = ?

Options:
A) \(\sqrt{\cfrac{B - 2}{2 - A}}\)
B) \(\sqrt{\cfrac{B - 1}{2 - A}}\)
C) \(\sqrt{\cfrac{B - 1}{A - 2}}\)
D) \(\sqrt{\cfrac{2 - B}{2 - A}}\)
Solution:
Ans: (b) 


ACos2θ + BSin2θ = Sin2θ + 2Cos2θ

BSin2 θ – Sin2 θ = 2 Cos2 θ – ACos2 θ

Sin2 θ (B - 1) = Cos2 θ (2 - A)

\(\cfrac{B -1}{2-A} = \cfrac{Cos^2θ}{Sin^2θ}\)

\(\sqrt{\cfrac{B-1}{2-A}}\) = Cotθ

Knowledge Expert

· commented

· 1 Months ago

Dear Student,

we inserted proper solution.

Hope you have a great learning with us!!

All the Best,
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