Deduce:- `sin^2 beta + cos^2 beta = 1`.

The value of sin^2 alpha cos^2 beta + cos^2 alpha sin^2 beta + sin^2 alpha sin^2 beta +cos^2 alpha cos^2 beta is

If 2 sin^(2)beta -cos^(2)beta =2 , then beta is

If alpha, beta, gamma are direction angles of a line l, then prove that cos^(2)alpha+cos^(2)beta+cos^(2)gamma=1 . Hence, deduce that sin^(2) alpha + sin^(2) beta + sin^(2) gamma = 2 .

If A=[(cos^(2)alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)] and B=[(cos^(2)betas,cos beta sin beta),(cos beta sin beta, sin^(2) beta)] are two matrices such that the product AB is null matrix, then alpha-beta is

(cos alpha + cos beta)/( sin alpha - sin beta) + (sin alpha + sin beta)/( cos alpha - cos beta ) =

Find the value of, cos alpha cos beta, cos alpha sin beta, -no alpha-sin beta, cos beta, 0 sin alpha cos beta, sin alpha sin beta, cos alpha] |

If cos alpha+cos beta=0=sin alpha+sin beta then cos2 alpha+cos2 beta=

If cos alpha + cos beta + cos gamma = 0 = sin alpha + sin beta + sin gamma then (a) sin ^ (2) alpha + sin ^ (2) beta + sin ^ (2) gamma = (3) / ( 2) (b) sin ^ (2) alpha + sin ^ (2) beta + sin ^ (2) gamma = (3) / (4) (c) cos ^ (alpha) + cos ^ (2) beta + cos ^ (2) gamma = (3) / (2) (d) cos ^ (2) alpha + cos2 beta + cos2 gamma = -1