2tan-1a-ba+btanθ2=
cos-1[acosθ+b][a+bcosθ]
cos-1[a+bcosθ][acosθ+b]
cos-1acosθ[a+bcosθ]
cos-1bcosθ[acosθ+b]
Explanation for the correct option:
Find the value of given expression:
Given, 2tan−1a−ba+btanθ2
=cos−11−a−ba+btan2θ21−a−ba+btan2θ2 ∵2tan-1x=cos-1(1-x21+x2);-1<x<1
=cos−1a+b-(a-b)tan2θ2a+b+(a-b)tan2θ2
=cos−1a1-tan2θ2+b1+tan2θ2a+atan2θ2+b1-tan2θ2
=cos-1a1–tan2θ21+tan2θ2+ba+b(1–tan2θ2(1+(tan2θ2
=cos-1[acosθ+b][a+bcosθ] ∵1-tan2θ21+tan2θ2=cosθ
Hence, Option ‘A’ is Correct.
The factors of 8a3+b3−6ab+1 are: