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Question : 167 of 180
Marks:
+1,
-0
Solution:
We have,
=cos−1p+cos−1q+cos−1r=π
⇒cos−1p+cos−1q=π−cos−1r
⇒cos−1(pq−√1−p2√1−q2)=π−cos−1r
⇒pq−√1−p2√1−q2=cos(π−cos−1r)
⇒pq−√1−p2√1−q2=−cos(cos−1r)
⇒pq−√1−p2√1−q2=−r
⇒pq+r=√1−p2√1−q2
⇒(pq+r)2=(1−p2)(1−q2)
⇒p2q2+r2+2pqr=1−p2−q2+p2q2
⇒p2+q2+r2+2pqr=1
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