`((2P+20)sin theta)/(2P+(2P+20)cos theta)=(2 P sin theta)/(P+2 P cos theta)`

ABCD is a trapezium such that AB and CD are parallel and BC bot CD . If angleADB = theta, BC = p and CD = q , then AB is equal to (a) ((p^(2)+q^(2))sin theta)/(p cos theta +q sin theta) (b) (p ^(2) + q ^(2)cos theta)/(p cos theta +q sin theta) (c) (p ^(2)+ q^(2))/(p^(2)cos theta +q^(2) sin theta) (d) ((p^(2) +q^(2))sin theta)/((p cos theta + sin theta)^(2))

If tan theta = p/q then (p sin theta - q cos theta)/(p sin theta + q cos theta)=

If tan theta = p/q then (p sin theta - q cos theta)/(p sin theta + q cos theta)=

If theta _(1) ,theta _(2), theta _(3),.....theta _(n) are in A.P. then (sin theta_(1) + sin theta _(2 ) +...+ sin theta _(n))/( cos theta_(1) + cos theta _(2) + ...+ cos theta _(n))

If tan theta=(p)/(q) then (p sin theta-q cos theta)/(p sin theta+q cos theta)=

If p and p_1 be the lengths of the perpendiculars drawn from the origin upon the straight lines x sin theta + y cos theta = 1/2 a sin 2 theta and x cos theta - y sin theta = a cos 2 theta , prove that 4p^2 + p^2_1 = a^2 .

If tan theta=(p)/(q) then the value of (p sin theta-q cos theta)/(p sin theta+q cos theta) is

If tan theta = p/q , proved that (p sin theta -q cos theta)/(p sin theta+q cos theta) is (p^2-q^2)/(p^2+q^2) .