If cos−135−sin−145=cos−1x then x equal to
We have cos−135−sin−145=cos−1x ⇒sin−1[√1−a25]−sin−145=cos−1x ⇒sin−145−sin−145=cos−1x ⇒cos−1x=0⇒x=cos0=1 ∴x=1
If A, B and C are angles of a triangle, then the determinant ∣∣ ∣∣−1cos Ccos Bcos C−1cos Acos Bcos A−1∣∣ ∣∣ is equal to
(a) 0 (b) -1 (c) 1 (d) None of these