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Question : 64 of 120
Marks:
+1,
-0
Solution:
Given:
cos−1x+cos−1y+cos−1z=π
cos−1x+cos−1y=π−cos−1z
⇒cos−1[xy−√(1−x2)(1−y2)] =π−cos−1z
⇒[xy−√(1−x2)(1−y2)] =cos(π−cos−1z)
⇒[xy−√(1−x2)(1−y2)] =−cos(cos−1z) (∵ cos (π - θ) = - cos θ)
⇒[xy−√(1−x2)(1−y2)]=−z (∵ cos(cos‌−1x) = x)
⇒√(1−x2)(1−y2)=xy+z
Squaring both sides, we get
⇒(√(1−x2)(1−y2))2=(xy+z)2
⇒(1−x2)(1−y2)=x2y2+z2+2xyz
⇒1−x2−y2+x2y2=x2y2+z2+2xyz
⇒x2+y2+z2+2xyz=1
∴x2+y2+z2=1−2xyz
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