Explanation
(sin(π2n)+cos(π2n))2=n2
1+2sinπ2ncosπ2n=n4......[cos2θ+sin2θ=1]
⇒1+sinπ2n−1=n4
sin(π2n−1)=n−44 AS n=+ve.≠1 and sinθ≤10<n−44≤1 ∴4<n≤8
The number of solutions of the equation 8tan2θ+9=6secθ in the interval (−π2,π2)
sin6θ+sin4θcos2θ−sin2θcos4θ−cos6θ
=sin4θ(sin2θ+cos2θ)−cos4θ(sin2θ+cos2θ)
=sin4θ−cos4θ[∵sin2θ+cos2θ=1]
=(sin2θ+cos2θ)(sin2θ−cos2θ)
=sin2θ−cos2θ
Hence, option D is the correct answer.
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