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Question : 25 of 160
Marks:
+1,
-0
Solution:
2sin−1x+sin−1[2x√1−x2] +3cos−1x−cos−1(4x3−3x) Put
x=cos‌θ⇒θ=cos−1x =2sin−1(cos‌θ)+sin−1(2‌cos‌θ‌√1−cos2θ) ‌‌+3cos−1(cos‌θ)−cos−1(4cos3θ−3‌cos‌θ) =2sin−1[sin(‌−θ)+sin−1∣2‌cos‌θ‌sin‌θ] =2(‌−θ)+2θ+3θ−3θ =π−2θ+2θ+3θ−3θ =π, when
x∈{−1,‌}
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