10(sin2θ)2+15(1−sin2θ)2=6 Let sin2θ=t⇒10t2+15(1−t)2=1610t2+15−30t+15t2=625t2−30t+9=0(5t−3)2=0sin2θ=53 and cos2θ=5216(25)427×27125+8+8125=125×5250=52
JEE Main 2025 — Mathematics Trigonometry
If 10sin4θ+15cos4θ=6, then the value of 16sec8θ27cosec6θ+8sec6θ is:
Held on 4 Apr 2025 · Verified 6 Jul 2026.
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