For the principal values, evaluate each of the following: `tan^(-1){2cos(2s in^(-1)1/2)}` `"cot"[sin^(-1){cos(tan^(-1)1)}]`

Correct option is: C) cos \(\theta\)  + sin  \(\theta\) \(\frac{cos\theta}{1-tan\theta} + \frac{sin\theta}{1-cot\theta}\)= \(\frac {cos\theta}{\frac {1-sin\theta}{cos\theta}} + \frac {sin\theta}{\frac {1-cos\theta}{sin\theta}}\) = \(\frac {cos^2\theta}{cos\theta - sin \theta} + \frac {sin^2\theta}{sin\theta - cos\theta}\) = \(\frac {cos^2\theta}{cos\theta - sin \theta} - \frac {sin^2\theta}{sin\theta...

Correct option is: B) Sin θ Cos θ \(Sin^3 \theta \, Cos \theta + Cos^3 \theta \, Sin \theta =\) sin \(\theta\) cos \(\theta\) (\(sin^2\theta + cos^2\theta\)) =  sin \(\theta\) .cos \(\theta\)  (\(sin^2\theta + cos^2\theta\) = 1)

Correct option is: A. cos θ sin θ

Correct Answer - C `I_(1)int_(0)^(pi//2)(cos^(2)x)/(1+cos^(2)x)dx` `=int_(0)^(pi//2)(cos^(2)(pi//2-x))/(1+cos^(2)(pi//2-x))dx` `=int_(0)^(pi//2) (sin^(2)x)/(1+sin^(2)x)dx=I_(2)` Also `I_(1)+I_(2)=int_(0)^(pi//2)((sin^(2)x)/(1+sin^(2)x)+(cos^(2)x)/(1+cos^(2)x))dx` `=int_(0)^(pi//2)(sin^(2)x+sin^(2)xcos^(2)x+cos^(2)x+sin^(2)xcos^(2)x)/(1+sin^(2)x+cos^(2)x+sin^(2)xcos^(2)x)` `=int(1+2sin^(2)xcos^(2)x)/(2+sin^(2)x cos^(2)x)dx=2I_(3)` `:. 2I_(1)=2I_(3)` or `I_(1)=I_(3)` or `I_(1)=I_(2)=I_(3)`

Correct Answer - C `AB=[(cos^(2) theta,),(cos theta sin theta,)][(cos^(2) phi,cos phi sin phi),(cos phi sin phi,sin^(2) phi)]` `=[(cos^(2) theta cos^(2) phi+cos theta cos phi sin theta sin phi ,cos^(2)theta cos phi...

(i) Since `1 in [-pi//2, pi//2], sin^(-1) (sin 1) = 1` (ii) Since `2 in [-pi//2, pi//2], sin^(-1) (sin 2) != 2` `:. Sin^(-1) (sin 2) = sin^(-1) (sin (pi)...

(i) Since `3 in [0, pi], cos^(-1) (cos 3) = 3` (ii) Since `4 !in [0, pi], cos^(-1) (cos 4) != 4` `:. Cos^(-1) (cos 4) = 2pi - 4`...

Correct Answer - `[{:(,1,0),(,1,1):}]`