sin theta=(p)/(sqrt(p^(2)+q^(2)))(q)/(d)quad log_(0)-q sin theta=p cos theta

If sin theta= (p)/(sqrt(p^(2)+q^(2))) then prove that qsin theta=p cos theta

5. sin p theta+sin q theta=0

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 : p*cos theta=sqrt(p^(2)+q^(2)).sin theta, "then" : q * sin theta=

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

int((sin theta-cos theta)d theta)/((sin theta+cos theta)sqrt(sin theta cos theta+sin^(2)theta cos^(2)theta))=p tan^(-1)sqrt((sin theta+cos theta)^(q)+r) where p,q,r and c are constants,then 2p+q+2r equals

ABCD is a trapezium such that AB|CD and CB is perpendicular to them.If /_ADB=theta,BC=p, and CD=q, show that AB=((p^(2)+q^(2))sin theta)/(p cos theta+q sin theta)

Solve : cos p theta = sin q theta.

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