The locus of the point `(r cos alpha cos beta, r cos alpha sin beta, r sin alpha)` is

The locus of the point (r sec alpha cos beta, r sec alpha beta, r tan alpha) is

The locus of the point (2sec alpha cos beta, 2 sec alpha sin beta, 2tanalpha) is

sin alpha=sinbeta, cos alpha=cos beta then

Evaluate {:[( cos alpha cos beta , cos alpha sin beta , -sin alpha ),( -sin beta , cos beta, 0),( sin alpha cos beta, sin alpha sin beta, cos alpha ) ]:} =0

(i) sin (alpha + beta) = sin alpha cos beta + cos alpha sin beta) (ii) sin (alpha-beta)= sin alpha cos beta- cos alpha sin beta Proof

If cos alpha + cos beta =0 = sin alpha + sin beta " then " cos (alpha - beta )=

If sin alpha = sin beta and cos alpha = cos beta then

If alpha, beta are solutions of a cos theta + b sin theta = c where a, b, c in R and a^(2) + b^(2) gt 0, cos alpha ne cos beta, sin alpha ne sin beta then prove that sin alpha + sin beta = (2bc)/(a^(2) + b^(2))

If alpha, beta are solutions of a cos theta + b sin theta = c where a, b, c in R and a^(2) + b^(2) gt 0, cos alpha ne cos beta, sin alpha ne sin beta then prove that sin alpha sin beta = (c^(2)-a^(2))/(a^(2) + b^(2))